i missed you guys SOOOO much [Archive]

Integral

May4-04, 04:11 AM

They both represent the same point in the Real Number system. This is what you fail to understand. A decimal or binary number (or any other base) is simply a representation of a point in the Real numbers. .1(base3) represents the same point as .333... base 10. It is not about the representation but the point being described. Like wise 1 represents the same point as .990... (base 10) or an infinitely repeating representation of the largest digit in any base.

Sorry that you do not like the way the real numbers are constructed. But the fact is that is the way it is. The basic properties of the Real number system is INDEPENDENT of ANY representation. Representations of a point on the real number line is more like the image on a movie screen. It gives you something to look at, but is not the same as the screen itself. The image can change but the screen remains the same. You can only discuss the image because you have no knowledge of the screen.

I am sure that you have spent hours thinking about the properties you wish to give infinity. What you have is not useful or even very interesting. Sorry. My opinion, but then you are simply expressing your opinion.

Hurkyl

May4-04, 06:36 AM

Ram, can your interpretation of the decimals (exactly) solve the equation 3x=1? If not, why should anyone use your interpretation instead of one that can exactly solve this equation?

ram2048

May4-04, 10:52 AM

Ram, can your interpretation of the decimals (exactly) solve the equation 3x=1? If not, why should anyone use your interpretation instead of one that can exactly solve this equation?

3x=1
x=1/3
x=.333r(1/3) exactly and rational

furthermore going backwards

x=.333r(1/3)
3x=.333r(1/3) x 3
3x=.999r(3/3)
3x=1

uart

May4-04, 11:28 AM

3x=1
x=1/3
x=.333r(1/3) exactly and rational

Hehe 1/3 = .333r(1/3), it just gets funnier. You mean 1/3 of 1/infinity there dont you RAM.

You know I knew that you'd believe that \infty+1 \neq \infty, such a belief is actually an inevitable consequence of believing the 0.999 is not equal to 1.

For example, what would RAM say 1-(1+0.999)/2 equals. It's obvious that he reply that it was 0.0...05, where the "0...0" denotes (infinity + 1) zero's (to distinghish it from the infinity zeros that 1 - 0.999 = 0.0...01 has). :biggrin:

I'll give you one thing RAM, you're a great entertainer.

ram2048

May4-04, 11:35 AM

Hehe 1/3 = .333r(1/3), it just gets funnier. You mean 1/3 of 1/infinity there dont you RAM.

no. i mean Remainder 1 divided by 3

this expanded notation is part of the last calculated digit in a non-terminating series created by a rational fraction.

it is simply a way to create the true rational decimal value

it's detailed in the link... kinda (still prettifying the page)

russ_watters

May4-04, 11:58 AM

To be more general than Hurkyl: So what? So you've got a different way of defining infinity. Why should we care?

ram2048

May4-04, 12:04 PM

Zurtex

May4-04, 12:18 PM

ram2048, I've been thinking about this and you've almost convinced me.

However there is just one little flaw that I see that you may be able to patch up. You are of the belief that \infty + 1 \neq \infty Which means you must have some default value for infinity, for example: \infty_d such that:

0.\overline{0}1 = \frac{1}{\infty_d}

Am I correct so far? If so how do you define this \infty_d and furthermore for this to hold true you must have a way of defining all other infinities right? Otherwise your system would just fall apart.

If you can show me this way of mathematically creating relationships of all infinites with one natural infinity and describe their mathematical relationship to real numbers then I will believe you :smile:

ram2048

May4-04, 12:33 PM

However there is just one little flaw that I see that you may be able to patch up. You are of the belief that \infty + 1 \neq \infty Which means you must have some default value for infinity, for example: \infty_d such that:

0.\overline{0}1 = \frac{1}{\infty_d}

Am I correct so far? If so how do you define this \infty_d and furthermore for this to hold true you must have a way of defining all other infinities right? Otherwise your system would just fall apart.

i'm not certain that 0.\overline{0}1 = \frac{1}{\infty_d} they are infinitessimals (someone used that words somewhere i'm just going to assume it's a real word) but the way they are defined and created are quite different, thus i don't think you can create and equality between them. lemme consider.

If you can show me this way of mathematically creating relationships of all infinites with one natural infinity and describe their mathematical relationship to real numbers then I will believe you :smile:

cute but i doubt your sincerity :D

in any case, let's assume a default infinity to begin with. if all rational transforms upon that infinity result in a net positive gain then the resulting infinity is greater than the original and we will use a different notation for the new infinity \infty_e. likewise were the transforms to result in negative "gain" then the resulting infinity would be less than the original infinity. let's use \infty_c

we can accurately say \infty_c < \infty_d < \infty_e .

<crosses fingers for tex coding>

Zurtex

May4-04, 01:49 PM

You have not defined the original infinity so how can you say that? Nor have you defined a relationship, for example would:

\infty_d - \infty_c = a

Where a is some finite number and according to your system it would be a specific finite value, what would it be? Even if it is not a finite value according to your system it would be a specific infinite number, what would it be? I am of course assuming a > 0 under the system you have stated.

So I wonder would:

\frac{1}{{\infty_d}^{\infty_c}} = 0.\overline{0}n ?

Where n is some finite digit? Or would this actually be 0?

Furthermore your notation is flawed, you are only going to have a finite amount of infinites where as I can think of an infinite number of ways to increase the value of a number. That's all the questions I can write down at the moment as I have to go, have at least another few dozen questions to see how your system works.

matt grime

May4-04, 05:20 PM

You ought to at least understand that hyper-real, or infinitesimal numbers, are not real numbers and so attempting tp cite them in defence of your ramblings of ignorance about the real numbers won't win you any arguments.

Hurkyl

May4-04, 06:05 PM

CHAPTER 3

Proof #1

In this proof they call upon a previously existing summation theorem that assumes to be correct and then proceed to apply the variables to come out with the conclusion. NO ONE can accurately sum any series to infinity to begin with. This entire proof stems from that fallacy.

Would you care to explain what you mean? Do recall that the DEFINITION of the sum of an infinite series is:

\sum_{i=0}^{\infty} a_i = \lim_{m \rightarrow \infty} \sum_{i=0}^{m} a_i

As you correctly point out, one cannot directly add an infinite number of terms... which is why we're not silly enough to do it that way.

Proof #2

on the third line. x or .9999... is subtracted from both sides and the assumption is made that you will get an even number of 9 as the result.

Actually, it is fairly easy to do this in a perfectly rigorous way.. In the way mathematicians define things, the position of each digit in a decimal number is labelled by an integer.

To give a hint of the flavor of how things are done done, I'll ask you this question:

In what position does 0.999~ have a '9' that doesn't appear in 9.999~?

(again, let me remind you; by the mathematical definition, that position has to be an integer... and no integers are infinite)

Proof #3

something INFINITELY SMALL is never NOTHING.

Incorrect; zero is the only infinitessimal real number. FYI, the definition of "infinitessimal" is:

x is an infinitessimal if and only if |x| is smaller than any positive real number.

Since the decimal numbers are constructed so that they are a model of the real numbers, it follows that zero is the only infinitessimal decimal number.

Proof #4

Go ahead and do the division for yourself. One divided by three. .33333333~ forever. Where is it wrong?

It's not.

well three does not divide evenly into one no matter how many digits you take it to, so that remainder of 1 will always exist just as the digits will never terminate.

But, the way mathematicians define the decimals, they don't have remainders. So 1/3 is 0.333~.

so 1/3 would equal 0.33~ R1/3

So this begs the question, in your system, what is 1/3? We can keep substituting:

1/3 = 0.33~ R1/3 = 0.33~ R0.33~ R 1/3 = 0.33~ R0.33~ R 0.33~ R 1/3

but we'll never come to a satisfactory representation. :frown:

But consider a much more pressing issue; how can a remainder possibly make sense if we want a decimal representatino of √2, or pi?

Also, allow me to suggest something for "fun". Let's assume for a moment that you actually have a well-defined and consistent number system.

Allow me to invent a new number system by decreeing that any two numbers that have the same decimal part are equal (even if they have different remainders), and furthermore I decree true any statement that can be proved from this decree. (So, for instance, 0.999~ = 1 because you've been able to prove 0.999~ R3/3 = 1 R0) I wonder what properties this number system has?

Hurkyl

May4-04, 06:41 PM

Hrm.

4/3 = 4 * 0.333~ R(1/3) = 1.333~ R(4/3)
4/3 = 1 + 1/3 = 1 + 0.333~ R(1/3) = 1.333~ R(1/3)

Problem...

ram2048

May4-04, 07:00 PM

\frac{1}{{\infty_d}^{\infty_c}} = 0.\overline{0}n ?

i don't know that looks complicated. is it actually a real question or are you coming up with new inventive ways to waste my time? :D

CHAPTER 3

Proof #1

Would you care to explain what you mean? Do recall that the DEFINITION of the sum of an infinite series is:

\sum_{i=0}^{\infty} a_i = \lim_{m \rightarrow \infty} \sum_{i=0}^{m} a_i

As you correctly point out, one cannot directly add an infinite number of terms... which is why we're not silly enough to do it that way.

Proof #2

Actually, it is fairly easy to do this in a perfectly rigorous way.. In the way mathematicians define things, the position of each digit in a decimal number is labelled by an integer.

To give a hint of the flavor of how things are done done, I'll ask you this question:

In what position does 0.999~ have a '9' that doesn't appear in 9.999~?

shifting the decimal for such a number is tricky business because it causes an inequality in the infinities in both expressions.

(again, let me remind you; by the mathematical definition, that position has to be an integer... and no integers are infinite)

Proof #3

i'm kinda going through this with someone in another thread at the moment as well and there is no satisfactory answer yet thus far. but expression 1 would have infinite 9's and expression 2 would have infinite+1 9's.

Incorrect; zero is the only infinitessimal real number. FYI, the definition of "infinitessimal" is:

x is an infinitessimal if and only if |x| is smaller than any positive real number.

Since the decimal numbers are constructed so that they are a model of the real numbers, it follows that zero is the only infinitessimal decimal number.

But, the way mathematicians define the decimals, they don't have remainders. So 1/3 is 0.333~.

So this begs the question, in your system, what is 1/3? We can keep substituting:

1/3 = 0.33~ R1/3 = 0.33~ R0.33~ R 1/3 = 0.33~ R0.33~ R 0.33~ R 1/3

but we'll never come to a satisfactory representation. :frown:

But consider a much more pressing issue; how can a remainder possibly make sense if we want a decimal representatino of √2, or pi?

the remainder is only used to create rational decimals for rational fraction values that would otherwise create non-terminating irrationals

the r(1/3) is not at the end of the series, rather it is part of the last digit. so you wouldn't be able to string it like that.

you could consider the last digit to be 10/3 rather than 3 <basically>

Also, allow me to suggest something for "fun". Let's assume for a moment that you actually have a well-defined and consistent number system.

Allow me to invent a new number system by decreeing that any two numbers that have the same decimal part are equal (even if they have different remainders), and furthermore I decree true any statement that can be proved from this decree. (So, for instance, 0.999~ = 1 because you've been able to prove 0.999~ R3/3 = 1 R0) I wonder what properties this number system has?

sounds like you're describing the current number system...

Hrm.

4/3 = 4 * 0.333~ R(1/3) = 1.333~ R(4/3)
4/3 = 1 + 1/3 = 1 + 0.333~ R(1/3) = 1.333~ R(1/3)

Problem...

you messed it up :D

4/3 = 4 * 0.333~ R(1/3) = 1.333~ R(1/3)
3x.333~ made 999~ the 3/3 completed the 1 which was part of the last digit 9. creating 10/10 that completes all the way to the top to make 1.

the process remainder is part of the last digit not after it.

Integral

May4-04, 08:23 PM

3x.333~ made 999~ the 3/3 completed the 1 which was part of the last digit 9.
You now need to share with us what you mean by a digit. How is something "part of a digit"

ram2048

May4-04, 09:03 PM

sorry if it wasn't clear enough.

when you divide out 1/3 each digit is created by a process of 10/3. each digit = 9/3 however. that 1/3 left over is the remainder i'm talking about.

Integral

May4-04, 10:10 PM

You are correct, it is not clear. What do you mean when you use the word "digit"?

ram2048

May4-04, 10:29 PM

suppose i wanted to get the true decimal value of 1/3 (one divided by three)

i would start doing division (the tedious way)

0.333~ r 1
3|1.0
9
10
9
10

each "digit" of 3 in the decimal notation resulting was created by dividing 10 by 3.

to use hurkyl's notation that he brought up
f(n)=0 where n ≥ 1
f(n)=3 where n < 1

those digits.

except in this case

f(n)=(10/3) where n = ∞

hope that formatting works

Integral

May4-04, 10:39 PM

Ok, how is something "part" of a digit? Is not a digit a single bit of information how am I to interpret something as part of a single piece?

You have shown me how you generate digits but you still have not told me what a digit is.

ram2048

May4-04, 10:45 PM

Hurkyl

May4-04, 11:47 PM

shifting the decimal for such a number is tricky business because it causes an inequality in the infinities in both expressions.

Can you state yourself in a precise way?

Here are two interesting questions for your system:

What is 9 + 0.99~?
What is 0.99~ * 10?

Can you even divine a difference in your system?

In the standard mathematical definition of decimals, the two are the same string of digits (and thus equal as decimals), because for every integer n, the digit in the n-th place in the first is equal to the digit in the n-th place in the second.

i'm kinda going through this with someone in another thread at the moment as well and there is no satisfactory answer yet thus far. but expression 1 would have infinite 9's and expression 2 would have infinite+1 9's.

Mathematicians do study these kinds of things; they're called order types. There really is a way to rigorously say that a sequence has length "infinity + 1" (where "infinity" is the order type of the positive integers). However, there is no such thing as an "infinity - 1"; the problem is that in order to take things off of the end of an ordering, the ordering has to have an end.

Because the digits to the right of the decimal place are indexed by the positive integers (or negative, if you prefer), this ordering doesn't have an end from which things can be removed. This is why it is possible to left shift all of the digits and have "one more" than you started with. (Mathematically, it still has the same order type and cardinality)

You, like most others who have similar ideas, seem to be imagining an order type that does have an end. Now, if you were to define the decimals as having two groups of digits to the right of the decimal place, an "a group" and a "b group" so that the positions went like this:

| 1a, 2a, 3a, ... | ... 3b, 2b, 1b |

(where I've used pipes (|) to denote the boundary of a group)

then when you multiplied this new type of decimal by 10, you would indeed "lose" a digit off the right end; the 1b position would be filled with a zero.

sounds like you're describing the current number system...

Exactly right. :smile: I was trying to demonstrate that even if you did have a consistent number system with infinitessimals, you can recover a number system that behaves exactly like the standard mathematical system. The point is to show relative consistency; if you believe the standard system to be flawed, then your system must be flawed as well because your system can be used to create the standard system.

Hurkyl

May5-04, 06:20 AM

Here's a puzzler. If you are only introducing remainders in the case of dividing things, then what is e^(-ln 3)?

Integral

May5-04, 06:22 AM

okay... a digit(in base ten) is one integer of values 0 through 9 which exist in sequence to describe real numbers such that these digits multiplied by various powers of 10 can be summed to create the number they describe.

i'm going to assume that your follow-up question will be "if you just said the digit is integers 0 through 9 how can you have a digit that is 10/3"

the answer is, this is new notation
So once again what do you mean by part of a digit? You are going to need a better definition then simply saying it is notation. The ellipsis is simply notation, I can define it in a few words; it stands for a infinitely repeating pattern. Now what do you mean by "part of a digit" ?

You seem to have agreed that infinity must have some default value, which to me says that you have confused infinity with a large finite number, all of your work is consistent with that conclusion. If you replaced infinity with any large finite number all of your work would make sense and you would have little disagreement from anyone on this board.

Your belief that 1 - .999... >0 means that the interval (.999...,1) has a finite non zero length, thus there must be a hole in the real line at this point, are you saying that .999... is the hole? are there other holes? If there is a hole at .333... why is it that .1(base 3) exists is it not right in the middle of the hole that you have placed at .333...?

I think it is your system that is full of holes, it simply does not work. Your time would be better spent actually attempting to learn how the real number system works rather then trying to reinvent a square wheel. It's your time.

Hurkyl

May5-04, 06:36 AM

Consider this.

Choose any of the 9's in 0.999~.
Consider the 9 in the next position to the right.
This 9 gets shifted underneath the 9 you chose when you multiply by 10.
Thus, 9.999~ has a 9 in the same position as the 9 you chose in 0.999~.

If this argument holds for any 9 you choose, then we have proved that 0.999~ cannot have a 9 in a position that 9.999~ does not.

Again, this works because decimal numbers don't have right ends; each position has a next position to the right.

Zurtex

May5-04, 09:40 AM

:rolleyes: Not trying to waste your time at all, just showing you that your system does not work at all ever. But lets go back to basics, you agreed that:

0.3\overline{3} = \sum_{n=1}^{\infty} \frac{3}{10^n}

Well sorry to say but if you work that out using simple sums to infinity of geometric seris you get:

\sum_{n=1}^{\infty} \frac{3}{10^n} = \frac{1}{3}

ram2048

May5-04, 12:01 PM

The point is to show relative consistency; if you believe the standard system to be flawed, then your system must be flawed as well because your system can be used to create the standard system.

not if the flaw is inherent in the creation of the standard. eg: rounding processes.

Here's a puzzler. If you are only introducing remainders in the case of dividing things, then what is e^(-ln 3)?

i have no idea what that is. please explain -ln

has a finite non zero length, thus there must be a hole in the real line at this point, are you saying that .999... is the hole? are there other holes? If there is a hole at .333... why is it that .1(base 3) exists is it not right in the middle of the hole that you have placed at .333...?

i'm saying that because these numbers have infinity included in their nature the gaps between them are beyond human comprehension and can only be measured or described by "infinitessimal" notations.

Choose any of the 9's in 0.999~.
Consider the 9 in the next position to the right.
This 9 gets shifted underneath the 9 you chose when you multiply by 10.
Thus, 9.999~ has a 9 in the same position as the 9 you chose in 0.999~.

If this argument holds for any 9 you choose, then we have proved that 0.999~ cannot have a 9 in a position that 9.999~ does not.

yes but consider a static infinity.

let's say you have two numbers of .999. Each with \infty_d
number of 9's as decimal digits.

you multiply one of them by 10.

compare the two. if they both have \infty_d number of 9's still you have created a gap at the very end such that when you subtract the two there is not an integer result.

if you "create" another 9 (wackiness) then one has \infty_d+1(9's) and the other has \infty_d. hence they're using different infinities and the calculation is incorrect.

Well sorry to say but if you work that out using simple sums to infinity of geometric seris you get: \sum_{n=1}^{\infty} \frac{3}{10^n} = \frac{1}{3}

sums to infinity are wrong which is what i've been trying to say forever now :D

Zeno's paradox is not obliterated by "sums to infinity" within the confines of the problem set forth it creates a true statement that the destination cannot be reached

russ_watters

May5-04, 01:24 PM

you should care because the traditional way of viewing it leads to inaccuracies. such that something cut infinitely results in nothing.

it's fine by me if you want to embrace something that produces the wrong results due to fallacious logic.

go ahead Math is a tool with which we can model the real world. Its of course necessary that the capabilities of the tool exceed the requirements of any concievable application.

What you see as a flaw in math is actually just part of its power and flexibility. But the problem of course, is that in order to get the correct answers from math, you have to use it correctly: I think it is your system that is full of holes, it simply does not work. Your time would be better spent actually attempting to learn how the real number system works rather then trying to reinvent a square wheel. It's your time. Not only do you see holes that aren't there, you're trying to patch them with a wire mesh. Yeah - it really would be better for you if you learned not only how math works, but how to use and apply it correctly.

Zurtex

May5-04, 01:42 PM

sums to infinity are wrong which is what i've been trying to say forever now :D
Then you think 0.\overline{3} is inherently wrong anyway and thus this discussion is pointless.

If you think sums to infinity are wrong you must then believe that we have got values such as e, \pi and \sqrt{2} wrong. Please tell use how you define such values.

Hurkyl

May5-04, 03:41 PM

not if the flaw is inherent in the creation of the standard. eg: rounding processes.

It's straightforward to show that this methodology is consistent; the "danger" is that it might wind up just saying that everything is equal.

i have no idea what that is. please explain -ln

ln is the natural logarithm (LN, but lowercase)

i'm saying that because these numbers have infinity included in their nature the gaps between them are beyond human comprehension and can only be measured or described by "infinitessimal" notations.

If you can measure or describe them by infinitessimal notations, doesn't that place them in the realm of human comprehension? :smile:

Something needs clarified here; are you saying that the real numbers has holes, or just the decimal numbers (whatever you think the decimal numbers are)?

yes but consider a static infinity.

I can't; I don't know what a "static infinity" is.... but from

you have created a gap at the very end

I'm assuming you mean an order type that has an end. As I stated, if you are using an order type that has an end, then this argumentat is correct. However, this argument fails when the order type has no end. Since there is no last integer, the decimals as they are mathematically defined has no end, so your argument fails.

sums to infinity are wrong which is what i've been trying to say forever now :D

Are you disagreeing with the claim that

\lim_{m \rightarrow \infty} \sum_{n=1}^m \frac{3}{10^n} = \frac{1}{3}

? Or are you disagreeing with

\sum_{n=1}^{\infty} \frac{3}{10^n} = \lim_{m \rightarrow \infty} \sum_{n=1}^m \frac{3}{10^n}

?

ram2048

May5-04, 04:14 PM

disagreeing with

\sum_{n=1}^{\infty} \frac{3}{10^n} = \lim_{m \rightarrow \infty} \sum_{n=1}^m \frac{3}{10^n}

that.

the sum will never equal its limit because infinity is infinite

when you say .333 that's a sum when you say 1/3 that's its limit

i'm sure i'm not saying it correct but it should be correct enough to understand.

.333 approaches 1/3 but never reaches it.

.999 approaches 9/9 but never reaches it.

ps> i have no knowledge of how to perform logarithmic calculations and don't have a sci calc handy (not that it would help since i wouldn't know what to put in to begin with..)

Zurtex

May5-04, 04:21 PM

Ahh well that clears it up, you simply don't understand what infinity, a limit, a sum to infinity or a recurring decimal is.

Hurkyl

May5-04, 04:41 PM

the sum will never equal its limit because infinity is infinite

Ok, then this is one precise point where we differ.

Mathematically, \sum_{n=1}^{\infty} a_n means \lim_{m \rightarrow \infty} \sum_{n=1}^m a_n. (i.e. it's neither a theorem nor an assumption; this is how an "infinite sum" is defined)

ram2048

May5-04, 05:06 PM

ah so i have to work at getting that abolished first then eh?

because it makes no logical sense that the sum of a convergent series equals it's limit when the very definition of convergent means it will never reach its limit.

so there we go.

arildno

May5-04, 05:30 PM

the very definition of convergent means it will never reach its limit.

??????????????????

russ_watters

May5-04, 07:10 PM

the very definition of convergent means it will never reach its limit.

so there we go. This is another definition on which you are wrong (or 'don't agree with' - which in this case is pretty much the same thing). "Convergent" means it does reach its limit. "Divergent" means it doesn't.

ram2048

May5-04, 08:47 PM

Convergent: - Mathematics. The property or manner of approaching a limit, such as a point, line, function, or value.

hmm Wolfram Mathworld studiously avoids defining convergent...

possibly another conspiracy... ;D

Hurkyl

May5-04, 10:19 PM

A sequence is convergent iff it has a limit.

BTW, have you considered the sequence: 0, 0, 0, 0, ...? :smile:

I think what you meant to say is that, generally, each term of the sequence will be inequal to the limit. (But, of course, that is not always true)

because it makes no logical sense that the sum of a convergent series equals it's limit when the very definition of convergent means it will never reach its limit.

I think what you meant to say is that it violates your "common sense".

Even with your definitions this doesn't follow; even if the partial sums don't "reach" the limit, why should that suggest anything about the infinite sum?

ram2048

May5-04, 11:38 PM

for fun's sake let's go back to Zeno's Paradox

1/2 + 1/4 + 1/8 + 1/16...

every step only completes half of the remaining value converging towards 1.

BUT let us consider the last digit

1/\infty

here's where we get the problem i think

1/\infty would complete only half distance theoretically as well, BUT current math doesn't believe beyond infinity.

AND

they believe that anything divided by infinity is the same number

hence you COULD have 2/\infty and complete the whole remaining distance and it would be the same as adding 1/\infty

pure conjecture of course. i think that in order to continue our understanding of math we have to find ways to define stuff better. having 20 different definitions for infinity is a pain in the patoot

Integral

May6-04, 02:45 AM

having 20 different definitions for infinity is a pain in the patoot

I sure understand what you are saying, that is why I cannot figure out why you insist on having so many definitions of infinity.

If you would simply make an effort to understand the SINGLE definition used by Mathematicians your world would be much simpler.

In the extended real line infinity is simply defined as being larger then all real numbers. A single simple definition. Thus any real number times infinity is still infinity because it is still larger then any real number. Likewise 1 over infinity must be smaller (in absolute value) then all real numbers, the only thing that meets this condition is 0, therefore \frac 1 \infty = 0 simple and consistent. You only add unnecessary complexity with your infinite number of infinites.

You are right that a set of infinitely shrinking intervals does indeed contain something (I think you phrased it that no matter how much you divide it up there is still something there). That something is a single point. Will you agree that the "length" of a single point is 0? That is 1-1=0? So does it not make sense to say that the length of the interval resulting in infinite divisions (which results in a single point) has length zero?

Will wait for your reply to make sure we are on the same page.

matt grime

May6-04, 03:33 AM

Ram seems to be having the standard issue that sequences can tend to limits and not actually reach them after a finite number of steps. That aand the facthe thinks there's a last term in an infinite sequence.

Would you like to list the 20 different definitions of infinity? There's only one meaning for infinite, and no need to use the word infinity if it confuses you.

ram2048

May6-04, 05:37 AM

In the extended real line infinity is simply defined as being larger then all real numbers. A single simple definition. Thus any real number times infinity is still infinity because it is still larger then any real number.

so you consider infinity to be a number governed by its own rules such that infinity is greater than itself (greater than all real numbers)

or is it not a number, such that it doesn't break that definition being greater than itself should infinity + 1 > infinity

and as far as the interval being a single point i don't believe it so. I still believe everything is further infinitely divisible beyond infinity.

arildno

May6-04, 05:41 AM

ram_1024:
(I won't be part of this ego-doubling process from your side; however, I'm not a churl, so I won't start halving it either)

I'd like to point out a few implications of your own ideas, rather than making clear to you why standard maths is consistent, and hence, why your attacks on it is quixotic at best.

(Others on this forum are by far more competent than myself in doing such an explanation, and if you had bothered to read, and tried to digest what they have patiently written to you, you would have stopped these ridiculous attacks long ago)

Implications of your ideas:
a)
Now, you say that a sequence: 0.1, 0.01, 0.001 and so on does not converge to 0,
but to some number 0.....1 (with an infinite number of zeroes in between)
Clearly then, a sequence: 0.2, 0.02, 0.002 goes to some number:
0.....2 (Right?)

In particular, this is perfectly in accordance that each term in the second sequence is the double value of the same term in the first sequence:
For example, 0.02=2*0.01, so it should make sense that the end value,
0.....2 is the double of 0....1 (Correct?)

b) Look now at the sequence:
1, 0.1, 0.01 and so on.
At every single instance, the term in this sequence is 10 times bigger than the same term in the sequence 0.1, 0.01 and so on.

Hence, by your logic, 1, 0.1, 0.01 must go to the number 0....10
(You can't escape this conclusion, sorry about that!)

c)
However, the first sequence is simply a subsequence of the last one; i.e. subsequences in your system doesn't go to/converge to the same values as the sequence itself.
In short, the whole convergence concept is blown to smithereens, and it is meaningless in the first instance to say that something "produces" or goes to
something at all; i.e. every single utterance you have made is mumbo-jumbo and nothing else.

Have a good day!

matt grime

May6-04, 06:46 AM

there is nothing contradictory about this infinity+1 not being greater than infinity, since the definition is that it is greater than all real numbers; infinity is not a real number, so infinity+1 is still infinity, unless you're talking ordinals, when w and w+1 are distinct ordinals, but that is a different system from the extended real line again. amazingly it always seems that these crackpot attempts at showing inconsistency are inconsistent, but also that someone non-crackpot has thought it through and offered something that is consistent and does it properly: surreal numbers, hyperreal numbers, the extended complex plane, ordinals, cardinals...

Hurkyl

May6-04, 06:47 AM

Suppose I do a thorough analysis of red M&Ms. Does that mean there are no green M&Ms?

The destination, indeed, does not appear in Zeno's analysis, but there's no reason in particular to think that Zeno's analysis covers the entirety of the motion in question.

However, there is not a "remaining distance" either. Zeno's analysis covers every position up to (but not including) the destination. If the destination is 1 meter away, then Zeno's analysis covers all positions x where 0 <= x < 1. The proof of this requires the Archmedian property posessed by the real numbers: for every real number r there is an integer n such that n > r.

It goes roughly as follows: let x be any real number in 0 <= x < 1. Let y = 1/(1-x). By the archmedian property, there is an integer n > y. By induction, we can find an integer m with 2^m > n. Thus, there is an integer m with 2^m > y, so 1/(2^m) < 1-x and 1 - 1/(2^m) > x. However, 1 - 1/(2^m) is the "current position" of the runner after the m-th step of Zeno's analysis. Thus, the position x has been considered by Zeno.

1/2 + 1/4 + 1/8 + 1/16...

every step only completes half of the remaining value converging towards 1.

BUT let us consider the last digit

Did you mean last term?

And why would there be a last term? Each term is of the form 1/2^m where m is an integer and there is no last integer...

so you consider infinity to be a number governed by its own rules such that infinity is greater than itself (greater than all real numbers)

The system Integral describes is called the "extended real numbers". In this system, infinity and -infinity are extended real numbers that are not real numbers.

russ_watters

May6-04, 07:38 AM

so you consider infinity to be a number governed by its own rules such that infinity is greater than itself (greater than all real numbers) Looks like we also need to define "real numbers" for you. The word "real" isn't arbitrary as you are using it, it has a specific definition in math as well. http://en.wikipedia.org/wiki/Real_numbers

In mathematics, the real numbers are intuitively defined as numbers that are in one-to-one correspondence with the points on an infinite line—the number line. The term "real number" is a retronym coined in response to "imaginary number". Notice, while the real number line is infinite, "infinity" is not a point on the line, hence "infinity" is not a real number. Your above objection is based on a misunderstanding of the definition of "real numbers." (you also characterized "infinity" as a "digit" a few posts ago - also incorrect).

Seriously, ram, you have a lot to learn about math. What we're talking about here is largely high school stuff. If you would only accept that you have a lot to learn and decide to learn it, you'd be much better off.

ram2048

May6-04, 07:18 PM

i can goto 5 sites and pull a different definition of real numbers, irrational numbers, and infinity.

and you talk to ME about inconsistency. I can't use your words because your definitions are "Mumbo jumbo"

i say something then you correct me with a different definition that means the same thing.

the point of the matter is, you say the number line is infinitely divisible but then you say when you divide it by infinity "it can't be determined" or "is close enough to zero we'll just call it zero"

i'm trying to establish a notational system that allows for infinite division AND allows for division beyond that.

you guys say the current system works, why mess with it, but totally balk at the idea of replacing it with something that works BETTER.

sentimental value i guess... :|

ram2048

May6-04, 07:31 PM

Implications of your ideas:
a)
Now, you say that a sequence: 0.1, 0.01, 0.001 and so on does not converge to 0,
but to some number 0.....1 (with an infinite number of zeroes in between)
Clearly then, a sequence: 0.2, 0.02, 0.002 goes to some number:
0.....2 (Right?)

these are not even converging sequences under my notation. you would have .00...02, .00...002, .00...0002, . . .

Have a good day!

i didn't! you cursed me didn't you!

ram2048

May6-04, 07:35 PM

Did you mean last term?

And why would there be a last term? Each term is of the form 1/2^m where m is an integer and there is no last integer...

of course i meant last term. Integral has me talking about digits and then you jump on me when i get a word wrong. okay okay spanish inquisition over here..

and yes there is no last integer by MY notation, but there IS by the current one. such that 1/Infinity = 0 = limit.

it's your system though, you KNOW it better than me why do i have to tell you?

Hurkyl

May6-04, 09:13 PM

i can goto 5 sites and pull a different definition of real numbers, irrational numbers, and infinity.

I wouldn't doubt it. There are often many equivalent ways of defining something.

For instance, (IIRC) the first rigorous definition of the real numbers was that a real number is an equivalence class of cauchy sequences of rational numbers. (In laymen's terms, any real number is identified with the sequences of fractions that approach it)

The usual modern definition preferred would be to define the real numbers as a complete ordered field. (In laymen's terms, +, -, *, /, and < all work "properly", and there are no "holes")

And it can be proved that the first definition satisfies the requirements of the second definition. Conversely, anything that satisfies the second definition is isomorphic (in laymen's terms, the same) to the first definition.

(Incidentally, 5? Really? I can only think of 3 definitions of real numbers you could be reasonably expected to find online, and only one definition of irrational. Different definitions of infinity wouldn't surprise me, because there are a lot of different concepts that are (sloppily) called "infinity")

the point of the matter is, you say the number line is infinitely divisible but then you say when you divide it by infinity "it can't be determined" or "is close enough to zero we'll just call it zero"

Actually, we say "Please be more specific".

"Infinitely divisible", taken entirely literally, means that it can be divided into an infinite number of terms. (Notice I did not say "an infinity of terms") In particular, the real line can be divided into c (= |R|) terms.

c is a thing called a cardinal number. It is not a finite cardinal, so it must be an infinite cardinal. (Some call it a transfinite cardinal, simply because so many laymen get confused when things are called infinite)

Division is not well-defined on cardinals, because multiplication is not very nice. For example, 1 * c = 2 * c. If you could divide by c, you would get 1 = 2, which is bad.

i'm trying to establish a notational system that allows for infinite division AND allows for division beyond that.

Learning math might be helpful to see how to do this. :smile:

Here's a simple approach to defining such a system:

Consider all (real) rational functions in x. E.G. things like 6, 7 + x^2, and (1 + 3x + 4x^5)/(4x + 3x^7)

+, -, *, and / can all be defined and function "properly" in this field. You can order this field by decreeing that x is bigger than any real number, and then extending the definition of < to accomodate this decree. So, for example, 7 + x^2 is infinite, because it is bigger than any real number. Proof:

Let r be any real number.
r < x
1 < x
r = 1 * r < 1 * x < x * x < x * x + 7
Thus r < x^2 + 7

Similarly, (1 + 3x + 4x^5)/(4x + 3x^7) is an infinitessimal.

If you don't like x, maybe you could use w (omega), a common symbol for transfinite numbers.

There's another number system (whose technical details are very difficult to follow) called the hyperreals which have transfinites and infinitessimals, but, for the most part, behave exactly like the real numbers. You might find information on this by searching for "non-standard analysis". Last time I went looking, there was actually an undergraduate calculus text in PDF format somewhere on the web that develops calculus using nonstandard analysis (i.e. with infinitessimals and transfinite integers, etc) instead of the usual way.

you guys say the current system works, why mess with it, but totally balk at the idea of replacing it with something that works BETTER.

We're "balking" because you are claiming the current system doesn't work.

of course i meant last term. Integral has me talking about digits and then you jump on me when i get a word wrong. okay okay spanish inquisition over here..

Or, maybe I'm just making sure I knew what you meant. :rolleyes:

ram2048

May6-04, 09:24 PM

i never claimed it didn't work, i claimed it was inaccurate ;D

i walk on bridges and fly by airplane. i wouldn't do so if i didn't trust our current set of math at all

in any case, there's little point in arguing as it seems apparant that such things are not going to change overnight.

i thank you for the envigorating discussion, and humbly apologize for getting bent out of shape at you in previous threads, Hurk

matt grime

May7-04, 03:10 AM

yes there is no last integer by MY notation, but there IS by the current one. such that 1/Infinity = 0 = limit.

it's your system though, you KNOW it better than me why do i have to tell you?

There is no last integer in 'our' system, that you think we think there is is your error.

ram2048

May7-04, 05:48 AM

ooooooooooooooooookay

define a number less than 1/infinity but greater than 0 in your system

matt grime

May7-04, 06:17 AM

what do you mean by 1/infinity? infinity isn't a real number, so why should i be able to do that? have you not learned anything from this thread? you aren'ty dealing with the real numbers when you write that kind of thing.

in robinsonian analysis i believe the object you are talking about is labelled epsilon.

Zurtex

May7-04, 12:18 PM

Well I'd like to thank all the people who tried to help ram out on this thread, whether or not he still spouting rubbish you have really made me feel a lot better about numbers :biggrin:. I still can't believe your trying to get the point across and have not just banned him from posting on the maths forums and ignored him here, well done your good people.

Hurkyl

May7-04, 05:40 PM

define a number less than 1/infinity but greater than 0 in your system

This question, in general, doesn't make sense.

It cannot make sense in the context of the real numbers.

In the context of the extended real numbers, this cannot be done, since 1/infinity is 0.

In the context of the hyperreal numbers, if we rephrase your challenge (so that it makes sense) as:
"If w is a positive, infinite number, then define a number less than 1/w but greater than 0"
then an answer to your challenge would be 1/(2w).

ram2048

May7-04, 08:13 PM

i somehow wrote that all wrong :O

what i meant to say is the "last" integer in your system would be a function of infinity such that nint(infinity) = that integer (theoretically if you could USE that function)

but that's still getting off the point.

if you COULD then 1/2^nint(∞) would be the closest step for your consideration.

forgot where i was going with this O_O

meh...

Zurtex

May8-04, 08:22 AM

ram2048 what is a "last integer"?

That makes less sense to me than anything else you have written.

ram2048

May8-04, 07:34 PM

hell if i know, Hurkyl was talking about integers and blah blah no last integer..

but there IS a last integer because you have an upwards limit of infinity.

you can't define that number without it being an expression "of infinity" itself, so it kinda defies itself.

but there was a point i was trying to make such that a sum to infinity, even AT infinity does not equal its limit in such a convergent sieries as Zeno's paradox.

as with everything "Infinity related" it's all theory and you have to apply logic. if every "step" in the process or "term" computed is half the remaining, there will never be a process that is "whole of the remaining" because that breaks the rule set forth in the initial exercise. so even at infinity, or beyond infinity in the case of "hyperreal" blah blah "extended irrationals" or whatever you want the sum STILL doesn't equal the limit.

the same is true of .999~ every digit tacked onto the end brings it closer and closer to 1, but no digit is ever a 10 completing a whole "step"

but it really doesn't matter... whatever :D

chroot

May8-04, 08:40 PM

So there's an integer that is larger than every other integer, eh? Could you do us a favor and show us which integer is the biggest?

- Warren

ram2048

May8-04, 09:17 PM

the one closest to infinity? :O

chroot

May8-04, 10:03 PM

That's the stupidest thing I've ever heard.

- Warren

ram2048

May8-04, 10:26 PM

haha my point is proven

not only can you NOT get to infinity, ever. you can NOT get to the largest integer that is not infinity.

so what hurkyl said about every step being considered is kinda out the window eh?

in any case .999 can never be = to 1 because every step is 9/10's the remaining step towards 1 and "the destination is never in consideration"

there is no step covered that is 10/10ths the remaining distance

sums to infinity are a good approximation

Hurkyl

May9-04, 12:59 AM

What is one plus the "largest integer that is not infinity"?
What is two plus the "largest integer that is not infinity"?

(plus, here, means integer addition)

the same is true of .999~ every digit tacked onto the end brings it closer and closer to 1

Why (and how) are you tacking digits onto .999~?

not only can you NOT get to infinity, ever. you can NOT get to the largest integer that is not infinity.

But you forget, you can "get" to every positive integer. It's a simple proof by induction:

I can get to 1.
If I can get to n, then I can get to n+1 by adding 1 to n.
Thus, by induction, I can get to any positive integer.

If the number system about which you are speaking has numbers to which you cannot "get", then you're not speaking about the integers.

in any case .999 can never be = to 1 because every step is 9/10's the remaining step towards 1 and "the destination is never in consideration"

.999 can never be equal to 1 because it is equal to 999/1000. Of what steps are you speaking?

Hurkyl

May9-04, 01:02 AM

sums to infinity are a good approximation

Approximation of what?

lvlastermind

May9-04, 01:09 AM

ram doesnt understand that you cant have 1/infinity beacause infinity=0 and you cant divide by zero.

ram2048

May9-04, 01:59 AM

Why (and how) are you tacking digits onto .999~?

doesn't matter why and how, but if you must know it's by process of summation where each "step" resolves into one "digit".

9/10 + 9/100 + 9/1000 ...

.999 can never be equal to 1 because it is equal to 999/1000. Of what steps are you speaking?

you knew what i meant.. :P

if you accept Zeno's conjecture as true, within the confines of the problem set forth you can not reach the destination then you also accept that within mathematics .999~ can never equal 1 simply because they are the same problem with a different curve Zeno's is 1/2 and .999~ is 9/10ths.

lvlastermind

May9-04, 02:22 AM

if you accept Zeno's conjecture as true, within the confines of the problem set forth you can not reach the destination then you also accept that within mathematics .999~ can never equal 1 simply because they are the same problem with a different curve Zeno's is 1/2 and .999~ is 9/10ths.

but you will reach your destination with an infinite number of halfs. This is one example of how infinity doesnt act like a real number. Beacause it doesnt have a value but it still has meaning.

Zurtex

May9-04, 06:54 AM

haha my point is proven

not only can you NOT get to infinity, ever. you can NOT get to the largest integer that is not infinity.

:rofl: :rofl: :rofl: We don't add the things one at a time.

ram2048

May9-04, 09:50 AM

but you will reach your destination with an infinite number of halfs. This is one example of how infinity doesnt act like a real number. Beacause it doesnt have a value but it still has meaning.

no this is an example of how calculus uses "infinity" to approximate.

if something cannot logically EVER be something then infinity and forever it will not be it.

how can you possibly reason that the destination is reached? it means that the last step you took wasn't a half-step but a whole one.

i have described earlier how calculus accepts 1/∞ and 2/∞ as the same number because both would be equal to 0.

indeed if 1/∞ = 0 then it is not actually a step at all, so the runner MUST have reached the destination in step 1/2^nint(∞) where nint defines the nearest integer to infinity.

but then we would have the paradox of being able to halve that distance YET AGAIN. such that 1/2^[nint(∞)+1]

and so on... so somehow the conclusion that the destination CAN be reached has to be wrong. or 1/∞ = 0 is wrong. or both are.

i say both personally ;D

Zurtex

May9-04, 09:58 AM

Your argument is wrong.

1/n > 0

Where n is a positive integer. There is NO nearest integer to infinity, that's just silly. Your problem is that you do not seem to be able to grasp a concept of infinity and that it is not on the real number line.

ram2048

May9-04, 10:27 AM

excuse me, but you guys sum n to infinity ALL THE TIME

that means it's on the same number line, it may not be included in your set of reals, but it's still on the same line.

that means somewhere along the line increasingly greater numbers become infinite. if NOT sum n to infinity has NO meaning. like saying sum n to cow or sum n to vaccuum cleaner...

the definition of infinity provides a clear relation of "the concept" to known reals such that even though it's not a number it's a function of numbers so much so that it is possible to use it in calculations.

if you're saying a set of reals can never increase to infinity then you're basically accepting that given the infinite number of steps in Zeno's problem, the man will NEVER reach his destination. clear now?

if you understand that much go back to my other post and read the logical explanation on how even at infinity the destination cannot be reached, so it makes no real difference whether a "largest integer" is real or not. the outcome is still the same

ram2048

May9-04, 10:44 AM

sums to infinity are a good approximation

Approximation of what?

approximation of the actual value that would come out of the calculation if you calculated it out longhand. :|

logically

Zurtex

May9-04, 11:32 AM

Hurkyl

May9-04, 11:44 AM

doesn't matter why and how, but if you must know it's by process of summation where each "step" resolves into one "digit".

9/10 + 9/100 + 9/1000 ...

This notation looks like an infinite sum; I don't see any "steps".

(To keep things moving)

I'm presuming by "steps" you are first considering 9/10, then 9/10+9/100, then 9/10+9/100+9/1000, and so on. But, of course, none of these are 9/10 + 9/100 + 9/1000 ... (though the limit of these "steps" is)

if you accept Zeno's conjecture as true, within the confines of the problem set forth you can not reach the destination

I have no problem with that. Zeno only considers the motion up to (but not including) reaching the destination. Thus, it would be silly to think that the destination would be reached in the period he analyzes.

then you also accept that within mathematics .999~ can never equal 1 simply because they are the same problem with a different curve Zeno's is 1/2 and .999~ is 9/10ths.

How do you figure? If I do it with 9/10s, then Zeno considers each of these intervals of position [0, 9/10], [9/10, 99/100], [99/100, 999/1000] ...
Putting all of these intervals together yields the interval [0, 1).

excuse me, but you guys sum n to infinity ALL THE TIME

Right, and sums to infinity are defined by

\sum_{i=1}^{\infty} a_i = \lim_{m \rightarrow \infty} \sum_{i=1}^m a_i

which can be further resolved (by applying the definition of limit)
\sum_{i=1}^{\infty} a_i = L if and only if for every positive \epsilon there exists an integer N such that for any integer m greater than N, we have
[tex]
|L - \sum_{i=1}^m a_i| < \epsilon
[/itex]

Notice, in particular, that this is not logically equivallent to saying that you keep adding terms one by one until you've reached an infinite number of terms. (Though, IMHO, it's for the most part conceptually equivalent)

that means it's on the same number line, it may not be included in your set of reals, but it's still on the same line.

The number line only has real numbers on it, thus it doesn't have infinity on it. :tongue:

But, as mentioned before, mathematicians do use an extension of the real numbers which has a positive and negative infinity on each endpoint. But...

that means somewhere along the line increasingly greater numbers become infinite.

No it doesn't. A sequence of increasingly greater numbers can converge to infinity, but none of the individual numbers need be infinite... just like a sequence of numbers can converge to zero, but none of them need to be zero. (e.g. 1, -1, 1/2, -1/2, 1/4, -1/4, 1/8, -1/8, ...)

if NOT sum n to infinity has NO meaning.

I'll say it again, sum to infinity has this meaning:

\sum_{i=1}^{\infty} a_i = \lim_{m \rightarrow \infty} \sum_{i=1}^m a_i

if you're saying a set of reals can never increase to infinity

I'm saying no real number may be infinite, an entirely different statement.

you're basically accepting that given the infinite number of steps in Zeno's problem, the man will NEVER reach his destination. clear now?

I accept that; the destination is reached after the "steps" contemplated by Zeno.

approximation of the actual value that would come out of the calculation if you calculated it out longhand. :|

logically

How do you plan to add an infinite number of terms by longhand?

ram2048

May9-04, 02:38 PM

How do you figure? If I do it with 9/10s, then Zeno considers each of these intervals of position [0, 9/10], [9/10, 99/100], [99/100, 999/1000] ...
Putting all of these intervals together yields the interval [0, 1).

how is that different from [0, 1/2], [1/2, 3/4], [3/4, 7/8]...

ram2048

May9-04, 02:51 PM

i'm still not understanding your wording on this

I accept that; the destination is reached after the "steps" contemplated by Zeno

if every step is half the distance (how the problem is defined) then after the "steps" still leaves a whole destination left to cover. if every step is a fraction of 1 whole then somewhere along you believe 1/2 = 1

the only thing i can think of to account for this is your belief that infinity is always equal to itself no matter how you transform it such that 1/2 x ∞ = 1 x ∞.

something like that would make a last step possible but it's not logical at all.

Zurtex

May9-04, 03:01 PM

the only thing i can think of to account for this is your belief that infinity is always equal to itself no matter how you transform it such that 1/2 x ∞ = 1 x ∞.

something like that would make a last step possible but it's not logical at all.
Infinity isn't a real number and doesn't act like that one, so yes that would be correct.

Last possible step? What are you on about?

Hurkyl

May9-04, 03:27 PM

how is that different from [0, 1/2], [1/2, 3/4], [3/4, 7/8]...

Put all of those together and you also get the interval [0, 1).

i'm still not understanding your wording on this

My intent was to state this: the destination is not reached during the sequence of steps considered by Zeno, but that does not imply that the destination cannot be reached at some time that occurs later than the steps considered by Zeno.

However, this reminded me of some of the things I used to point out in Zeno's paradox discussions; the following statement is also true:

At any particular point in time, if it can be said that all of the steps considered by Zeno have occurred, then it is also true that the destination has been reached.

But allow me to emphasize; the destination is not reached during the steps considered by Zeno. In particular, in order to make this statement, I do not make the assumption that one of these steps covers the entire remaining distance. Additionally, I need not use "infinity" anywhere to prove my claim.

the only thing i can think of to account for this is your belief that infinity is always equal to itself no matter how you transform it such that 1/2 x ∞ = 1 x ∞.

I "believe" it because that is how multiplication by ∞ is defined in the extended real numbers.

something like that would make a last step possible but it's not logical at all.

You mean that it conflicts with your common sense. It is entirely logical because it can be proven rigorously from the definitions and axioms.

But, it shouldn't conflict with common sense. Consider this; when things grow without bound, they approach infinity, right? E.G. the sequence 1, 2, 3, 4, ... approaches infinity, and the sequence 2, 4, 6, 8, ... does as well.

Furthermore, multiplication is a continuous operation. If a sequence approaches a limit, then if I double every number in the sequence, then the limit gets doubled.

Now, 2, 4, 6, 8, ... is the double of the sequence 1, 2, 3, 4, ..., so the limit of 2, 4, 6, 8, ... is double the limit of 1, 2, 3, 4, ..., thus suggesting that ∞ = 2 * ∞ should be correct.

ram2048

May9-04, 03:57 PM

the destination is not reached during the sequence of steps considered by Zeno, but that does not imply that the destination cannot be reached at some time that occurs later than the steps considered by Zeno.

well since zeno considers to infinity, you believe that "beyond infinity" there lies a step such that 1/2 the remaining distance = the whole distance?

what i'm saying is infinity or not you believe you can take 1/2 a distance and it equals the whole distance. basically going back to what Integral said about the smallest thing being a point, what's the 'length" of a point there isn't a length so points cannot be used to fill a "gap" of any distance.

because going backwards in steps you'd also have to have 2x"the length of a point distance left to cover such that you could 'fill' it with your 1x point.

the convergence would look something like...

8 point lengths remaining (move (1/2 x 8) points)
4 point lengths covered

4 point lengths remaining (move (1/2 x 4) points)
2 point lengths covered

2 point lengths remaining (move (1/2 x 2) points)
1 point length covered

1 point length remaining (move (1/2 x 1) point. point is not divisible. move 1 point)
1 point length covered

i dunno if that makes anymore more sense at all, but that last consideration is the one i say wouldn't happen because it would be beyond the confines of the problem.

bear in mind i know a point has no "length" so don't lecture me on that. i'm using it to illustrate a point since it's the smallest conceivable thing.

so using that model, stuff has a limit of divisibility. the end step would break the rules set forth in the problem such that you'd be moving not 1/2 step but a whole step. while we calculate out with infinite steps we still reach that imposed limit caused by the point not being divisible.

but if it WERE divisible it STILL wouldn't be reached because we'd be faced with 1/2 a point 1/4 a point etc etc.

ram2048

May9-04, 04:13 PM

Now, 2, 4, 6, 8, ... is the double of the sequence 1, 2, 3, 4, ..., so the limit of 2, 4, 6, 8, ... is double the limit of 1, 2, 3, 4, ..., thus suggesting that ∞ = 2 * ∞ should be correct

i'm not seeing how that proves infinity=2xinfinity. seems quite the opposite because one sequence is twice the other. how are you saying they are equal?

consider:

when you say something grows you have to apply a rate. if the rate of growth is equal among 2 sequences such that sequence 1 growing double that of sequnce 2. then it's easy to prove that within any given frame of reference in time, sequence one is larger than sequence 2. even at infinity.

1,2,3,4... ...100,101,102... ...1001,1002,1003...
2,4,6,8... ...200,202,204... ...2002,2004,2006...

such that any line cutting perpendicularly into these 2 parallel lines will always yield a number and a number 2xgreater than it.

although if you were to grow things without a rate of time, these number sequences would "exist" as any number within their range at any given time, it would be impossible to accurately assess any relation in value when comparing them. 1 could = 3 or 300,000.

Hurkyl

May9-04, 04:23 PM

you believe that "beyond infinity" there lies a step such that 1/2 the remaining distance = the whole distance?

I believe there is nothing between Zeno's steps and the destination.

Consider that the sequence 2, 4, 6, 8, ... is a subsequence of 1, 2, 3, 4, ...,

2 4 6 8 ...
1 2 3 4 5 6 7 8 ...

Since the former can be formed by removing terms from the latter, how can the former have a bigger limit?

within any given frame of reference in time
although if you were to grow things without a rate of time,

What does time have to do with anything?

even at infinity.

The sequences aren't even defined at infinity. (unless your "infinity" is a positive integer, in which case it is finite, thus making your use of the term highly misleading)

these number sequences would "exist" as any number within their range at any given time

At any given time, a number sequence is a number sequence. They do not "exist" as anything else.

hello3719

May9-04, 08:24 PM

Ram2048, infinity is a PROCESS and not a number, it only means that you are constantly growing in relation to the certain set of numbers considered.

so you can see that infinity*2 is the same as infinite since in both cases the growing process is present which is what infinity means. growing twice as fast or growing is the same as growing.

ram2048

May9-04, 11:00 PM

I believe there is nothing between Zeno's steps and the destination.

so the sum of that sequence will never equal its limit...

if this is NOT the case you need to explain to me how we're processing along towards the destination with computation and all of a sudden there's no distance left and we're there. There has to be a definable process that gets us there.

Consider that the sequence 2, 4, 6, 8, ... is a subsequence of 1, 2, 3, 4, ...,

2 4 6 8 ...
1 2 3 4 5 6 7 8 ...

Since the former can be formed by removing terms from the latter, how can the former have a bigger limit?

What does time have to do with anything?

that's why i said growth has to be a function of time or reference in some way "rate" such that the steps can be measured. otherwise if sequence 1 "grew" at a rate 50,000 times slower than sequence 2, it wouldn't matter that they had the same limit or that sequence 1 was "twice the value" of sequence 2, [edit] at the beginning seq 1 would be greater than seq 2 but it would then be overtaken due to superior rate.

The sequences aren't even defined at infinity. (unless your "infinity" is a positive integer, in which case it is finite, thus making your use of the term highly misleading)

that statement was a conclusion stemming from the process. most processes that tend towards infinity have a very predictable output and logical conclusion. with the same growth "rate" and the same start point, the double sequence will hit infinity while the single will only be at 1/2 infinity.

this is the logical conclusion, nevermind that neither will EVER reach infinity,

At any given time, a number sequence is a number sequence. They do not "exist" as anything else.

not so, if no rate is applied to growth it means that it simultaneously exists as any allowable value within its field (start to limit)

----------------

and as a side note i want to know if this proof works by your math.

x=.999~
10x = 9.999~
10x - x = 9.999~ - .999~
9x = 9

x = 1
.999~ = 1

curious to see if that's allowed or if some crackpot made that up without running it by you guys.

Hurkyl

May9-04, 11:49 PM

Ram2048, infinity is a PROCESS and not a number

That's at least half incorrect. In number systems with a single nonfinite number (or some otherwise especially identifiable nonfinite number), that nonfinite number is often labelled "infinity". Its other uses (such as in \sum_{i=0}^{\infty}) are as a formal symbol to denote something in particular (IMHO it would be misleading to call things so denoted "processes").

so the sum of that sequence will never equal its limit...

The sum of an infinite sequence is, by definition, the limit of the partial sums, so...

if this is NOT the case you need to explain to me how we're processing along towards the destination with computation and all of a sudden there's no distance left and we're there. There has to be a definable process that gets us there.

Zeno's steps cover every point between the start and destination (but not including the destination). If Achilles has completed all of Zeno's steps, where could he be if not at (or past) the destination?

A more analytical answer might go as follows: if p(t) is the position of Achilles at time t, then, according to Zeno's steps, p(0) = 0, p(1/2) = 1/2, p(3/4) = 3/4, p(7/8) = 7/8, et cetera.

Now, lim_{n \rightarrow \infty} (1 - 1/2^n) = 1, and since motion is continuous, we conclude

\begin{align*}
p(1) &= p(lim_{n \rightarrow \infty} (1 - 1/2^n)) \\
&= lim_{n \rightarrow \infty} p(1 - 1/2^n) \\
&= lim_{n \rightarrow \infty} (1 - 1/2^n) \\
&= 1\end{align}

that's why i said growth has to be a function of time or reference in some way "rate" such that the steps can be measured.

That seems necessary to maintain your stance, but I can't see any reason that this is necessary in general.

not so, if no rate is applied to growth it means that it simultaneously exists as any allowable value within its field (start to limit)

Rephrase yourself. It is absurd to say, for instance, 1 = <1, 2, 3, 4, ...>, but I am interpreting you as suggesting that this equality can be correct.

and as a side note i want to know if this proof works by your math.

x=.999~
10x = 9.999~
10x - x = 9.999~ - .999~
9x = 9

x = 1
.999~ = 1

Yes, this is a valid proof.

It is not a valid "first" proof, though, in the sense that this proof depends on the fact that the decimal numbers are a model of the real numbers, so it cannot be used to prove, in the first place, that the decimal numbers are a model of the real numbers.

ram2048

May10-04, 12:10 AM

It is not a valid "first" proof, though, in the sense that this proof depends on the fact that the decimal numbers are a model of the real numbers, so it cannot be used to prove, in the first place, that the decimal numbers are a model of the real numbers.

darn cause i had a good disproof and proof on infinity+1≠infinity from that one :(

oh well

Zeno's steps cover every point between the start and destination (but not including the destination). If Achilles has completed all of Zeno's steps, where could he be if not at (or past) the destination?

he could still be in transit.

\begin{align*}p(1) &= p(lim_{n \rightarrow \infty} (1 - 1/2^n)) \\&= lim_{n \rightarrow \infty} p(1 - 1/2^n) \\&= lim_{n \rightarrow \infty} (1 - 1/2^n) \\&= 1\end{align}

i've always concluded that 1/∞ has a positive value. i've been trying real hard to understand you guys' "something turns into nothing" but it's not getting through :(

Rephrase yourself. It is absurd to say, for instance, 1 = <1, 2, 3, 4, ...>, but I am interpreting you as suggesting that this equality can be correct.

no no, that's not what i meant by it. i meant since it is a defined series of growth, without anything to denote the transition in steps such a rate or time value, it would be any one of those numbers at the exact instance of transformation.

it's part Schrodinger in a way i guess. but it's kinda veering on a tangent away from the issue at hand hehe.

in any case i'm pretty sure if i can convince myself that 1/∞=0 everything will be clear to me :O

Hurkyl

May10-04, 06:45 AM

he could still be in transit.

If he's still in transit, he has to be someplace between the start and finish.
If he's someplace between the start and finish, then he has not done all of Zeno's steps.

Thus, if he has done all of Zeno's steps, then he's not in transit.

The key thing about the real numbers, that you are not using, is that they are complete. Intuitively, that means they have no "holes"; this is usually stated as follows:

If A and B are nonempty collections of real numbers, such that A lies entirely on the left of B, then there is a number c seperating them.
More precisely,
Let A and B are nonempty sets of real numbers
For all a in A and b in B: a < b
Then, there exists c such that
For all a in A and b in B: a <= c <= b

This can be used to prove the Archmedian property; every number is smaller than some integer (and is thus finite) as follows:

Let A be the set of everything smaller than some integer. (Note that all integers are in A, because n < n+1)
Let B be the set of everything bigger than all integers.
Assume B is nonempty.
Then, there exists some c seperating them.

Thus, a <= c <= b for all a in A and b in B.

Now, c is either bigger than all integers (and thus in B), or it's not (and thus in A).

If c is in A, then it is smaller than some integer. Call it n.
Because c < n, c + 1 < n + 1. Thus, c + 1 is in A.
This is a contradiction because c + 1 > c, but because c + 1 is in A, c+1 <= c. Thus c cannot be in A.

Thus, c is in B.
c - 1 < c, so c - 1 must be in A. c - 1 must be smaller than some integer. Call it n. Because c - 1 < n, we have c < n + 1. However, n + 1 is an integer and is in A, so n + 1 <= c, which is a contradiction.

Thus c cannot be in B.

So our assumption that B is nonempty led to a contradiction: there's a hole between the finite and infinite numbers, but by definition the real numbers have no holes.

Thus, we conclude that there are no infinite numbers; every number is smaller than some integer, and thus the Archmedian property is proven.

Now, we turn back to your question. Suppose 1 / ∞ = e > 0. (whatever 1 / ∞ may happen to mean).

By the Archmedian property, 1 / e is smaller than some integer. Call it n.
Because 1/e < n, e > 1/n.

So, if you insist on maintaining that 1 / ∞ > 0, then there exists an integer n such that 1/n is smaller than 1/∞, which is absurd!

Roughly the same argument is used to prove that \lim_{n \rightarrow \infty} 1/2^n = 0.

(More directly, if you maintain that ∞ is a real number, then I could simply apply the Archmedian property to produce an integer bigger than ∞)

matt grime

May10-04, 07:29 AM

ram, have you considered stepping back, taking a deep breath and admitting to yourself that you don't understand the mathematics involved? that, in the extended numbers, infinity*2=infinity follows from the definitions of that system, and 1/infinity is zero. in the surreals, 1/w is not zero, but w is still not a real number.

As for zeno and trying to understand how you can 'go beyond infinity' as is being talked about, consider trying to learn about transfinite systems, and limit ordinals, you also need to stop thinking in terms of moving from one spot to the next, but in terms of having done all the previous moves (a vague introduction to transfinite induction).

If you tried to understand the limit concept properly instead of presuming you know what it is then none if this nonsense would happen.

ram2048

May10-04, 11:45 AM

So, if you insist on maintaining that 1 / ∞ > 0, then there exists an integer n such that 1/n is smaller than 1/∞, which is absurd!

that's because of the "self defined" upwards limit on infinity itself. BECAUSE you believe it to be all inclusive such that any value added to infinity has no meaning, you're presented with the upwards or downwards limits as such. but this still does NOT prove that the number would equal zero. consider 2 apples next to each other touching. there is no space in between them but they are not the same apple. BECAUSE infinity has a limit such that nothing can be greater than it, no SPACE can exist between 1/∞ and 0, but that doesn't prove they're the same number. infinity is not a number so it doesn't function as one in relation to this "archemidien rule". let's apply infinities "limit" to a real number and see how it works. let's make 500 be the highest possible number there can be. let's now say 1/500. is 1/500 = 0? it is most certainly NOT, yet this is the closest POSSIBLE number to 0 you can get because of the limit imposed.

ram, have you considered stepping back, taking a deep breath and admitting to yourself that you don't understand the mathematics involved? that, in the extended numbers, infinity*2=infinity follows from the definitions of that system, and 1/infinity is zero. in the surreals, 1/w is not zero, but w is still not a real number.

i doubt myself all the time, but until i can find the answers that convey meaning to me in such a way that i'm completely convinced and can properly relay this information to another person that might be having the same problem, i cannot quit.

As for zeno and trying to understand how you can 'go beyond infinity' as is being talked about, consider trying to learn about transfinite systems, and limit ordinals, you also need to stop thinking in terms of moving from one spot to the next, but in terms of having done all the previous moves (a vague introduction to transfinite induction).

well the point of zeno's exercise was that you could NEVER complete all the moves, that's why it was proposed as a paradox...

you want to explain to me then since you know how to do it? none of that fancy math stuff, just logically spell it out for me how it could EVER be possible to cover a whole distance if you can only move in halves of the remaining distance.

russ_watters

May10-04, 02:18 PM

i can goto 5 sites and pull a different definition of real numbers, irrational numbers, and infinity.

and you talk to ME about inconsistency. I can't use your words because your definitions are "Mumbo jumbo"| Quite frankly, the discussion should have ended here. If you can't even accept that there are specific definitions for these concepts, you can't ever hope to understand the definitions or how to apply them.

If you really do want to understand what we're trying to tell you, this is where you must start.

ram2048

May10-04, 03:42 PM

Quite frankly, the discussion should have ended here. If you can't even accept that there are specific definitions for these concepts, you can't ever hope to understand the definitions or how to apply them.

If you really do want to understand what we're trying to tell you, this is where you must start.

that's BS, russ. i wouldn't have known there were 5 or more definitions for commonly used math terms if i hadn't taken the time to research it and try to find which one was the correct one. i even go through the trouble to annotate myself with "quotes" when i'm unsure as to the correct term. this is possibly the best place to find proper definintions for these terms as if someone brings up the wrong one he is immediately and punitively apprehended... like within the hour. you guys are harsh ;D

and russ, if you have nothing positive to contribute, just don't. really.

matt grime

May10-04, 05:52 PM

these 'different' definitions of real numbers are all equivalent, that is they do the same thing, but that seems beyond your capability to understand.

teach your self transfinite induction, it isn't very hard, any idiot can understand it.

it has been explained to you patiently and accurately what the use of infinity is in mathematics, now it is time for you to go away and think about all that has been written and say what you don't understand, not what you think is worong because your mathematical spohistication is not such that you are at a point to say something is wrong, merely that you don't understand it, and that is a problem with you and not with mathematics.

we are not harsh, you are ignorant. that is not an insult, just a statement of fact.

ram2048

May10-04, 09:22 PM

it has been explained to you patiently and accurately what the use of infinity is in mathematics, now it is time for you to go away and think about all that has been written and say what you don't understand, not what you think is worong because your mathematical spohistication is not such that you are at a point to say something is wrong, merely that you don't understand it, and that is a problem with you and not with mathematics.

so you can't explain it without your fancy math. which leads me to the conclusion that you only believe it because it is popular and fashionable. If you don't understand the math well to be able to rationalize if logically, then all you are doing is parroting back text.

we are not harsh, you are ignorant. that is not an insult, just a statement of fact.

i am not knowledgable in the ways of "transfusion inducers" or whatever the heck you're talking about, that hasn't stopped me from disproving 4 "accepted" proofs. quite silly how it can all be taken down by an "ignorant" person.

Zurtex

May10-04, 10:53 PM

i am not knowledgable in the ways of "transfusion inducers" or whatever the heck you're talking about, that hasn't stopped me from disproving 4 "accepted" proofs. quite silly how it can all be taken down by an "ignorant" person.
That is your problem, you automatically assume you are right, you are unwilling to learn mathematics, you have not actually disproved anything and that makes you ignorant.

chroot

May10-04, 10:53 PM

When an intelligent man argues at length with a fool, it becomes difficult to tell them apart.

- Warren

ram2048

May10-04, 11:10 PM

That is your problem, you automatically assume you are right, you are unwilling to learn mathematics, you have not actually disproved anything and that makes you ignorant

i have disproved them, the ball is in your court. what purpose would it serve for me to disprove myself? unless you can disprove my disproofs then my way is correct. that's how this works.

when person A comes and tells me something and i prove him wrong, it doesn't lie on MY shoulders to disprove myself.

and i'm always willing to learn mathematics as long as they have some logical basis in reality. i will not blindly accept "Something equals nothing" unless convinced thoroughly that it is logically true.

actually, until you can come up with rational arguements as to why one of my proofs does NOT work, you have no right to call me ignorant. Come up with something on your own instead of monitoring the thread waiting to say "yea Hurkyl is right, you're stupid ram..." whenever he posts something.

chroot

May10-04, 11:12 PM

ram2048:

You didn't disprove anythnig. You might feel you did, but that's because you don't understand what a proof even is.

- Warren

Hurkyl

May10-04, 11:13 PM

no SPACE can exist between 1/∞ and 0

Where is (0 + 1/∞) / 2? If 0 and 1/∞ are different, then this quantity must lie between them. If there is nothing between them, then they must be equal.

infinity is not a number

Then what can 1/∞ possibly mean? Is that also not a number? If that's not a number, then how can the remaining distance (which is a number) be equal to 1/∞?

let's make 500 be the highest possible number there can be. let's now say 1/500. is 1/500 = 0? it is most certainly NOT, yet this is the closest POSSIBLE number to 0 you can get because of the limit imposed.

Ok. And?

well the point of zeno's exercise was that you could NEVER complete all the moves, that's why it was proposed as a paradox...

The problem is that Zeno never suggests any reason why you cannot.

you want to explain to me then since you know how to do it? none of that fancy math stuff, just logically spell it out for me how it could EVER be possible to cover a whole distance if you can only move in halves of the remaining distance.

You couldn't. It's quite fortunate that we're not restricted to moving in halves of remaining distance.

this is possibly the best place to find proper definintions for these terms

Then consider these definitions.

The usual definition of the real numbers used these days is along the lines of this one, slightly paraphrased from Buck's Advanced Calculus:

"The real numbers (R) constitute a complete simply ordered field"

In terms of axioms, this means:
R is a set of elements.
P is a subset of R (whose elements are called positive)
+ and * are two operations on elements of R.
0 and 1 denote particular elements in R

For any a, b, and c in R:
a + b is in R
a * b is in R
a + b = b + a
a * b = b * a
a + (b + c) = (a + b) + c
a * (b * c) = (a * b) * c
a * (b + c) = (a * b) + (a * c)
a + 0 = a
a * 1 = a
a + x = 0 can be solved for x
a * x = 1 can be solved for x if a is not zero
if a and b are positive, then so are a + b and a * b
either a is in P, -a is in P, or a = 0.

If A and B are nonempty subsets of R, and a <= b for any a in A and b in B, then there exists a number c such that a <= c <= b for any a in A and b in B.

The extended real numbers, as from Royce's Real Analysis is (I don't have my text handy, so I'm doing this from memory, and am probably saying the same thing in many more words):

The extended real numbers consist of the real numbers plus two additional elements, ∞ and -∞.
+∞ is positive, and -∞ is not.
For any extended real numbers a and b that are not equal to +∞ or -∞:
a + b and a * b in the extended real numbers is the same as in the real numbers.
a + +∞ = +∞
a + -∞ = -∞
a - +∞ = -∞
a - -∞ = +∞
If a is positive, then a * +∞ = +∞ and a * -∞ = -∞
If a is negative, then a *+∞ = -∞ and a * -∞ = +∞
a / +∞ = 0
a / -∞ = 0
+∞ + +∞ = +∞
-∞ + -∞ = -∞
+∞ - -∞ = +∞
-∞ - +∞ = -∞

In particular, 1 / +∞ = 0 simply because that's what is defined to equal, thus it is certainly not logical that 1 / +∞ be inequal to zero.

ram2048

May10-04, 11:37 PM

x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1

Assume .999~ has infinity number of digits 9.
when we multiply this by 10, 1 of 2 things happens:

1. a 0 is added to the end and all digits of 9 are shifted up one digit

OR

2. all digits of 9 are shifted up one digit and a digit of 9 is "created" to maintain that "infinite" number of 9's

out of these two possibilities, your proof favors the second notion, that all digits are shifted and an extra 9 is "created" such that the .999~ in the 9.999~ can be perfectly cancelled out to = 9 exactly.

now note that the expression .999~ having "infinity" number of digits 9. therefore 9.999~ MUST have "infinity+1" number of digits since this is proven when the subtraction takes place and a whole number digit 9 is left over.

9.999~ has "infinity+1" digits of 9
.999~ has "infinity" digits of 9

the creation of the extra digit of 9 causes inequalities in the infinities used in both numbers.

in short 9.999~ would have the same number of digits of 9 as .999~ and subtracting it would NOT yield the integer 9. using rational and logical step 1 instead it would output something like this: 8.999~1

going backwards we see this is absolutely true. take 9x which is 9 x .999~ and multiply the digits out 8.1 +.81 + .081 ... such that the sum EQUALS 8.999~1 [edit] woops added one too many steps there

QED

matt grime

May11-04, 05:00 AM

0.999... certainly doesn't have a finite number of nines so it must have an infinite number of them, and?

multiply by ten and we shift one place to the right and add one 'at the end' to maintain this infinite number? what utter utter nonsense.

we have shown you where you are wrong yet you refuse to recognize this fact, but then this is all opinion isn't it and our opinion is no more valid than yours, or so you seem to think.... again, wrong, these are based on definitions of what these symbols mean mathematically, something we understand and that you do not.

and round it goes in ever decreasing circles.

Hurkyl

May11-04, 06:47 AM

1. a 0 is added to the end and all digits of 9 are shifted up one digit

There is no end.

2. all digits of 9 are shifted up one digit and a digit of 9 is "created" to maintain that "infinite" number of 9's

Where is the "created" digit placed?

How can you maintain your position is logical when it requires there to be a rightmost digit... but every digit has another digit to its right?

ram2048

May11-04, 11:04 AM

no. by comparing the 2 numbers in such a fashion a "default infinity" is defined such that transformations are measurable.

it it absolutely logical that if you shift the decimal point down, that the number now has (infinity-1) number of 9's after the decimal point.

it doesn't make any sense to say it still has exactly "infinite" number of 9's when compared to our established "default infinity"

There is no end.

it doesn't MATTER if there's an end or not, with all real numbers, performing such a calculation WOULD shift digits and create a 0. what makes you think that this case would be different?

multiply by ten and we shift one place to the right and add one 'at the end' to maintain this infinite number?

that's what i'm talking about. this is visual proof that transformations using default infinity can yield perfect calculations. it's completely based off what you guys believe. if both numbers STILL had "infinity" number of 9's AFTER the decimal allowing you to cancel them out EXACTLY, then 9.999~ had 1 MORE digit 9 TOTAL than .999~ meaning it is NOT the same number as (10x.999~)

the proof is in the pudding when you go back and multiply the numbers out and acquire the exact same thing as you would subtracting them longhand.

9 x .999~ = (8.1 + .81 + .081...) = 8.999~1
10x - x = (9.999~0 - .999~) = ([9-1].[18-9][18-9][18-9]~[10-9]) = 8.999~1

Zurtex

May11-04, 11:56 AM

it doesn't MATTER if there's an end or not, with all real numbers, performing such a calculation WOULD shift digits and create a 0. what makes you think that this case would be different?

:rolleyes: Shift the digits? Create a 0? Once again you show your astounding ability not to understand even the most basic mathematics.

hello3719

May11-04, 12:56 PM

Ram2048, can you give us an exact definition of your "default infinity" and the definition of "infinity' ?

ram2048

May11-04, 01:00 PM

:rolleyes: Shift the digits? Create a 0? Once again you show your astounding ability not to understand even the most basic mathematics.

good work, Zurtex.

you caught me. apparantly i have NO IDEA how this is resolved

(10 x 8.16) - 3.57 =
81.6 (shift digits) - 3.57 =

81.60 (create 0)
- 3.57
78.03

yes... NO IDEA WHATSOEVER...

ram2048

May11-04, 01:16 PM

Ram2048, can you give us an exact definition of your "default infinity" and the definition of "infinity' ?

well i would probably define infinity as a transient variable such that the variable will always be a value greater than any number it is compared against.

default infinity would be a mathematical concept or technique in which the user establishes a base infinity for computations and any transforms based upon it are weighed mathematically to discern logical outcomes as to the nature of numbers defined using infinity.

instead of just saying infinity + 1 = infinity, which is slightly true because adding 1 to infinity means it's still infinite.

we could say ∞(d) + 1= ∞(e) and then extract meaningful values like ∞(e) > ∞(d) and ∞(e) - 1 = ∞(d)

ram2048

May11-04, 01:22 PM

also when using default infinity you could say things like

∞(d) - ∞(d) = 0

∞(d) / ∞(d) = 1

hello3719

May11-04, 01:26 PM

So how is it different from real numbers ?

hello3719

May11-04, 01:35 PM

since you've defined default infinity as being bigger than every other value compared to it then "default infinity" is bigger than "default infinity" +1 ?

then the contradiction arises here, (d) infinity + 1 must then be equal to inifinity since both are being compared and each of them must be bigger than the other.

ram2048

May11-04, 01:42 PM

it's not much different from real numbers at all which is what makes it useful.

standard "calculus" definition of infinity creates alot of "undefined" processes that lead to computational "dead-ends". like infinity minus infinity is unresolvable because it's not defined which infinity is the same, the natural inclination is to believe that they ARE the same and the result would be 0. default infinity provides a way to establish the relations of infinities we're using within a given mathematical process.

Infinity(d) - Infinity(d) = 0
Infinity(d) - Infinity(e) = unresolvable

define infinity(e) = Infinity(d) - 5

Infinity(d) - infinity(e) = 5

math
infinity(d) - (infinity(d) - 5) =
infinity(d) -1(infinity(d) + -1(-5) =
infinity(d) - infinity(d) + 5 =
5

something like that...

ram2048

May11-04, 01:48 PM

since you've defined default infinity as being bigger than every other value compared to it then "default infinity" is bigger than "default infinity" +1 ?

no i defined it as a variable "value" not a variable "number". it is bigger than all NUMBERS compared against it, not values.

but it allows trasformations using numbers to act on it such that numbers will cause its value to increase or decrease.

such that:

infinity(d) + 1 = infinity(e) (defining infinty(e)
infinity(e) > infinity(d) (logical comparison)

BUT

infinity(e) ≯ infinity(d) + 2 (adding the value of the number 2 to our default infinity)

hello3719

May11-04, 02:10 PM

Then there is no advantages of you're definition of inifinity over the system of real numbers by seeing the properties you are giving it.

well i would probably define infinity as a transient variable such that the variable will always be a value greater than any number it is compared against.

Example, If I take the number 1 , then by your definition of "inifinity" it must be 1 value greater than 1 so,
"inifinity" = 1+1 = 2

If I take the number 2, then inifnity = 2 +1 = 3

etc...

you can see we are constructing real numbers, but then you can see how our definition of inifinity helps in saying how much of your "inifinities" there is.

ram2048

May11-04, 02:23 PM

Zurtex

May11-04, 02:32 PM

not so.

by the current definition of infinity:

infinity + 1 = infinity
infinity + 2 = infinity
infitity + 3 = infinity
...

1 = 2 = 0 = 519,249 = any other number added to it.

the logical conclusion is that you CANNOT change the value of infinity in the current system so infinty - infinity MUST ALWAYS BE 0

and

infinity / infinity = 1

but according to current definitions infinity - infinity = unresolvable, infinity/infinity = unresolvable.

that's by the current system. that's the system with the logical flaws and contradictions.

i'm just creating a system that defines infinity to clear things up :rofl:

Wrong, if infinity can be greater than itself in either finite terms or infinite terms then infinity - infinity does not have to equal 0. Also infinity / infinity does not have to equal 1.

It is NOT A REAL NUMBER AND DOES NOT ACT LIKE ONE. Get used to it or don't bother doing maths.

Edit: If you want to create a system for classifying all infinites and works with real numbers go and do one and come back when you have it.

Zurtex

May11-04, 02:41 PM

good work, Zurtex.

you caught me. apparantly i have NO IDEA how this is resolved

(10 x 8.16) - 3.57 =
81.6 (shift digits) - 3.57 =

81.60 (create 0)
- 3.57
78.03

yes... NO IDEA WHATSOEVER...

:rolleyes: ROFL, it doesn't create a 0 there were an infinite number of 0s already there. Whoever taught you maths was either really bad or you weren't listening.

Consider this:

10 \cdot (3) = 30

Another way of writing it is:

10 \cdot (...0000003.0000000 \overline{0}...) = ...00000030.000000 \overline{0}...

Do you see anymore than an infinite number of 0s? I thought it was that your terminology was bad with using words like "shift" and "create" but now I see you actually think them.

Hurkyl

May11-04, 03:35 PM

The extended real numbers, as from Royce's Real Analysis is

I was wrong; division by infinity is not allowed in the definition presented by Royce.

no. by comparing the 2 numbers in such a fashion a "default infinity" is defined such that transformations are measurable.

Would you care to present a definition?

it it absolutely logical that if you shift the decimal point down, that the number now has (infinity-1) number of 9's after the decimal point.

Then present the logic.

it doesn't make any sense to say it still has exactly "infinite" number of 9's when compared to our established "default infinity"

Are you saying this statement doesn't make "sense", or that it is false?

it doesn't MATTER if there's an end or not, with all real numbers, performing such a calculation WOULD shift digits and create a 0. what makes you think that this case would be different?

Because you can't tell me where the 0 is.

And, heuristically speaking, why would you think it makes sense to "create" a 0 but not to "create" a 9?

if both numbers STILL had "infinity" number of 9's AFTER the decimal allowing you to cancel them out EXACTLY, then 9.999~ had 1 MORE digit 9 TOTAL than .999~

Where is that extra 9?

logical flaws and contradictions.

Would you care to present such a flaw and contradiction? As in a proof, and not your constant labelling of your intuition as logic?

Hurkyl

May11-04, 03:47 PM

Let's see if you can do something useful with your system of infinities. Let's start with some basic geometric "facts"; mathematicians once struggled to figure out how talk about them, let alone prove anything about them:

What does it mean to say that the collection of all numbers between 0 and 1 is a "continuum"? How do you prove it is a continuum?

How does one show that the collection of numbers that are either between 0 and 1 or 2 and 3 is not a continuum?

How does one say that the function f(x) = x is continuous?

And how would you prove it?

matt grime

May11-04, 03:48 PM

Ram, why do you want to treat infinity as a real number? I mean, in considering infininty minus infinity and so on? Why are you so fixed on it having a place in the real numbers, why do you think it is a number for a start? what is it that prevents you from thinking about and learning about all these other different things such as the extended numbers, the surreal,s the hyper reals, ordinals, cardinals, and so on. try it you might, well, learn something, rather than spouting the same ignorant garbage that every other idiotic crank can't deal with either

hello3719

May11-04, 04:21 PM

.

by the current definition of infinity:

infinity + 1 = infinity
infinity + 2 = infinity
infitity + 3 = infinity
...

1 = 2 = 0 = 519,249 = any other number added to it.

So you think that infinity - inifinity = 0 by our definition of infinity ?
( unless you meant something else by "current definition")

and if you think that there is a problem in having infinity-infinity being indeterminate this means that you don't know how to use the infinity concept. Understand the use of limits and you'll see how it is completly LOGICAL that "inifinity - infinity" is indeterminate. It is not at all a problem of the current mathematical system.

If you still think that our system has some major flaws,and your concept of "infinity" would solve problems, then try finding, for example, the area under the curve of some function using your definitions.

pallidin

May11-04, 06:39 PM

OK, ram, I will give SOME credence to your notions. But only some.
This might be of interest to you: I developed a theory some time ago that I called the "h-factor"
What it encompasses is beyond the scope of this discussion, but parts of it may be relevant here. I will attempt to illustrate those points:

Take an orange and weigh it.
Cut the orange in half.
Weigh the 2 separate pieces and sum the result.
The result will be less than the whole.
Why?
Because juice from the orange is left on the table.
The "juice on the table" is the expression of the energy lost from division transformation.

Not convinced of where this is leading to or it's importance?

Take a diamond and weigh it.
Cut it in half and weigh the separate diamonds.
The summed weight is less than the whole.
The diamond "dust" left over during division is the "missing" weight.

Still not convinced of serious mathematical implications concerning this?

Let's take this to the ultimate.
Take a theorectical rod of certain construction and nature such that when, broken in half, does not produce ANY fracture "dusting"
Surely, then, the two halfs will sum in weight to equal the weight of the original whole.
Correct!
But, what is forgotten here?
Energy is used AND DISSIPATED to enable that ideal event.

Is this getting clearer with regards to relevance in mathematics?
Look, in mathematics the division process is "pure", addressing adjunct concerns AFTER the "pure" division process.
Reality does not work this way.
In REALITY, division of a system requires the expenditure of energy upon INITIAL execution of the process.
Therefore(for simple example), 4/2=2 is INVALID because the process is held as being "pure", which is realistically impossible because energy MUST BE EXPENDED for the division process to occur even in this simple equation.

Got it?
Can ANYTHING be divided in the REAL WORLD(GR, SR, etc...) that does not require energy expenditure on INITIAL invocation of the event? No.
So, where is the energy expenditure in the equation 4/2=2 ?

Oh, some will say, that is all well and fine but does not apply to mathematics or when it does it is expressed in following equations.
Think again. It MUST be expressed in initial operand involvement because that is EXACTLY how REALITY works.

BTW, this is by no means subject only to division. All operations are subject to this.

Just some thoughts...

ram2048

May11-04, 06:49 PM

it it absolutely logical that if you shift the decimal point down, that the number now has (infinity-1) number of 9's after the decimal point.

9.999~ consider this to have "default infinity" number of digits 9
.999~ this number has (infinity(d)-1) which is demonstrably true when you subtract .999~ from 9.999~ you are left with exactly 1 digit 9.

it doesn't make any sense to say it still has exactly "infinite" number of 9's when compared to our established "default infinity"
Are you saying this statement doesn't make "sense", or that it is false?

both. it is both not logical and provably untrue as shown above.

it doesn't MATTER if there's an end or not, with all real numbers, performing such a calculation WOULD shift digits and create a 0. what makes you think that this case would be different?
Because you can't tell me where the 0 is.

And, heuristically speaking, why would you think it makes sense to "create" a 0 but not to "create" a 9?

because in base 10 calculations, multiplying by 10 creates a process which shifts ALL the digits. after defining a default infinity you can see this creates a distinguishable "gap" in comparison to the standard transient infinity. in order to maintain the equality in infinities the gap is filled by 0 during the subtraction.

the 0 is at the "end" of an infinite number of 9's. it doesn't matter that "you cannot get there" because in the calculation you're pitting it against a previously "measured" infinity such that you can logically maintain that when the digits are shifted, such a gap is created.

adding a 9 is illogical because 9 is a digit that has value, such that the number you create by adding it is a different number than what you started with.

let's try this with another repeating decimal.

x = .666~
10x = 6.666~
10x - x = 6.666~ - .666~
9x=6
x=2/3

according to calculus. 10x - x = 6 BUT 9x = 5.999~4

but using my system:
10x - x = 6.666~0 - .666~ = ([6-1].[15-6][15-6][15-6]~[10-6]) = 5.999~4
9x = 5.999~4

real numbers coming out perfectly. no approximation.

hello3719

May11-04, 07:38 PM

real numbers coming out perfectly. no approximation.

where do you see any approximation in these calculations?
x = .666~
10x = 6.666~
10x - x = 6.666~ - .666~
9x=6
x=2/3

and if you do the long division of 2 over 3 you will get .66666~
so clearly 2/3 = .666~ and we had x = .666~ and x=2/3
I don't see any problem in our system, it seems consistent and logically complete.

ram2048

May11-04, 07:49 PM

please read the post before replying

according to calculus. 10x - x = 6 BUT 9x = 5.999~4

ram2048

May11-04, 08:02 PM

hello3719

May11-04, 08:10 PM

:frown: 9.999~ consider this to have "default infinity" number of digits 9
.999~ this number has (infinity(d)-1) which is demonstrably true when you subtract .999~ from 9.999~ you are left with exactly 1 digit 9.

ok then, consider this:

suppose that .999~ this number has (infinity(d)-1) of the digit nine by your definitions, then you can agree that .4999~ has (infinity(d)-2) of the digit nine right?

k then by your reasonning .999~ - .4999~ has to have 1 digit 9 right?

but in fact it has 0 digit nine since .999~ - .4999~ = .500~

See the contradiction, interpreting the mathematical concepts purely by graphical representations of numbers is not a universally valid proof as you can easily fall into intuitive traps. Sorry, but in fact you didn't prove anything worthy of your system here.

ram2048

May11-04, 08:52 PM

suppose that .999~ this number has (infinity(d)-1) of the digit nine by your definitions, then you can agree that .4999~ has (infinity(d)-2) of the digit nine right?

k then by your reasonning .999~ - .4999~ has to have 1 digit 9 right?

but in fact it has 0 digit nine since .999~ - .4999~ = .500~

no you had it right. look here. .999~ and .0999~ exhibit a has 1 more 9 than exhibit b.

the 9 in that place is destroyed by the 4 when you do the subtraction of .999~ - .499~

that's why there's no 9 left over. there is no contradiction, visually or otherwise.

hello3719

May11-04, 09:02 PM

exactly, you were proving the problem visually, look again at your proof.

ram2048

May11-04, 09:11 PM

not understanding you.

please point it out for me

Hurkyl

May11-04, 09:15 PM

9.999~ consider this to have "default infinity" number of digits 9

Then what you mean by 9.999~ is not what mathematicians mean by 9.999~, unless you can somehow show that having a "default infinity" number of digits means exactly the same thing as saying that there is exactly one digit corresponding to each positive integer.

999~ this number has (infinity(d)-1) which is demonstrably true when you subtract .999~ from 9.999~ you are left with exactly 1 digit 9.

Where is that 1 digit 9?

And I forget if I asked this:
How does your system of decimals tell the difference between 9 + 0.999~ and 10 * 0.999~, seeing how they're both 9.999~?

both. it is both not logical and provably untrue as shown above.

It can't be both; nonsensical statements are just that, nonsensical, and cannot be logically manipulated. In particular, you can't prove them false. This is more evidence that you are not talking about logic.

because in base 10 calculations, multiplying by 10 creates a process which shifts ALL the digits. after defining a default infinity you can see this creates a distinguishable "gap" in comparison to the standard transient infinity.

Then, again, you aren't talking about the standard decimals. (Assuming "transient infinity" means the standard mathematical way of doing things)

This is called a "strawman argument". Your "decimals" are something different than the decimals of mathematics. While your statements may then be true about what you mean by "decimal", you are merely attacking a strawman; there are no grounds to suggest any of your statements also apply to the mathematical meaning of "decimal".

the 0 is at the "end" of an infinite number of 9's.

How can there be an end if every digit has another digit to its right?

And here's another stickler. If there's an "end" then what happens to the last 9 in 0.999~ when you divide by 10?

x = .666~
10x = 6.666~
10x - x = 6.666~ - .666~
9x=6
x=2/3

according to calculus. 10x - x = 6 BUT 9x = 5.999~4

This has nothing to do with calculus; it's arithmetic. But 9x is 5.999~ = 6, and 5.999~4 is something that has no meaning in standard mathematics.

no you had it right. look here. .999~ and .0999~ exhibit a has 1 more 9 than exhibit b.

If you consider .999~ to have "one more" 9 than .0999~, then why don't you consider 9.999~ to have "one more" 9 than .999~?

What about this proof?
x = .999~
x/10 = .0999~
x - x/10 = .999~ - .0999~ = .9
(9/10) x = .9
9x = 9
x = 1

ram2048

May11-04, 09:28 PM

What about this proof?
x = .999~
x/10 = .0999~
x - x/10 = .999~ - .0999~ = .9
(9/10) x = .9
9x = 9
x = 1

assume .999~ in line 1 to be default infinity number of 9's
in line 2 assume x/10 would shift all decimals right one digit and create a 0, yet maintain the same "default infinity" number of digits 9
.999~ - .0999~ =.8999~1

9/10 = .9
9/10 x .999~ = .8999~1

ram2048

May11-04, 09:38 PM

How does your system of decimals tell the difference between 9 + 0.999~ and 10 * 0.999~, seeing how they're both 9.999~?

they wouldn't be the same.

in case 1, 9.000000000~ even is added to .999~ . in case 2, 10 instances of .999~ are added.

if we were to set .999~(d) as default infinity number of 9's (from now on i'll use this notation) adding 9 would not cause the shift in digits, 10 x .999(d) would. such that 9+.999(d) - (10 x .999~(d)) = .000~9

hello3719

May11-04, 09:56 PM

remember this?

9.999~ consider this to have "default infinity" number of digits 9
.999~ this number has (infinity(d)-1) which is demonstrably true when you subtract .999~ from 9.999~ you are left with exactly 1 digit 9.

you SAW that there is 1 digit 9 left, this is visual and doesn't mean that "infinity(d)" - ("infinity(d)"-1) = 1

assume .999~ in line 1 to be default infinity number of 9's
in line 2 assume x/10 would shift all decimals right one digit and create a 0, yet maintain the same "default infinity" number of digits 9
.999~ - .0999~ =.8999~1

9/10 = .9
9/10 x .999~ = .8999~1

k so you just admitted that .999~ and .0999~ have the same "default infinity" number of digits 9. You are now giving your infinity the same property as our infinity.So "infinity(d)" - ("infinity(d)"-1) isn't anymore equal to 1.

ram2048

May11-04, 10:06 PM

i don't think you're understanding that i can take any "infinity" and define that as my "default infinity" and work calculations from there.

default infinity isn't the same for every calculation, it MUST be designated to be used.

all it does is create a base to work from, it doesn't matter what you use as a base, the logical transformations from there stem outward and resolve to the same conclusion.

the notation system does need work i admit that, but there are no contradictions in the system, visually or otherwise.

hello3719

May11-04, 10:22 PM

So in fact it is useless since you can't prove anything usefull about your system.

Hurkyl

May11-04, 10:37 PM

i know nothing of continuums.

not going to even go there

Ok, I rememebred some easier to describe problems that have similar trappings:

Suppose I have a circle. Point A is inside the circle and point B is outside the circle. Can you prove the line segment joining A and B intersects the circle?

Prove that x^2 = 2 has a solution. (In other words, prove that √2 exists)

Hurkyl

May11-04, 10:53 PM

Back to your "default infinity":

You suggest that ∞(d+1) = ∞(d) + 1.
Is this true in general? That ∞(a + b) = ∞(a) + b?
If so, note that we could define ω := ∞(0), then ∞(d) = ω + d.

Do any of the following make sense? If so, what are they?
∞(1/2)
2 * ∞(0)
(1/2) * ∞(0)
π * ∞(0)
∞(0)^2
∞(0)^π
√∞(0)
2^∞(0)
sin ∞(0)
∞(∞(0))

If this question makes sense, is ∞(0)^2 < 2^∞(0)? How would you prove your answer?

To what default infinity does this sequence approach:
0, 2, 2, 4, 4, 6, 6, 8, 8, 10, 10, 12, ...
and what about this sequence:
1, 1, 3, 3, 5, 5, 7, 7, 9, 9, 11, 11, ...
Which of the two infinities is bigger? Or are they the same?

What do you have to say about this sequence:
0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...
and how does it compare to
1, 2, 3, 4, 5, 6, ...
or to
1/2, 1, 3/2, 2, 5/2, 3, ...?

Can I define a number x to have a 9 in every finite position, but a 0 in every infinite position? If so, how does x compare to 10x - 9? If not, why not?

Can I define x to have 9 in every position and do the same thing? That means not only does it have a 9 in every finite position, but also in every infinite position. (So, for instance, there is no d so that the ∞(d+1)-th position doesn't have a 9 in it)

ram2048

May11-04, 11:01 PM

Ok, I rememebred some easier to describe problems that have similar trappings:

Suppose I have a circle. Point A is inside the circle and point B is outside the circle. Can you prove the line segment joining A and B intersects the circle?

i don't remember how to plot a circle, but i'm assuming this "proof" would involve me computing an exact value of π

Prove that x^2 = 2 has a solution. (In other words, prove that √2 exists)

a square with side length 1 has a corner to opposing corner diagonal or exactly √2 length using pythagorean. i'm not sure what you want me to do about it though, visually and logically they both exist, but can they be rationally measured?

i'll work on that but don't expect results anytime soon ;D

Hurkyl

May11-04, 11:10 PM

a square with side length 1 has a corner to opposing corner diagonal or exactly √2 length using pythagorean. i'm not sure what you want me to do about it though, visually and logically they both exist, but can they be rationally measured?

Blarg, right, square roots aren't a problem; the problem I meant is showing x^3=2 has a solution!

Actually, there is some issue related to square roots, I'll see if I can remember what it is.

Hurkyl

May11-04, 11:16 PM

Given a point O, and another point R:

Any point X is said to be:
Inside the circle if X = O.
Inside the circle if the segment OX is shorter than the segment OR.
On the circle if the segment OX is the same size as the segment OR.
Outside the circle if the segment OX is longer than the segment OR.

And since you bring up pi, here's another problem which is much trickier than the above one:

Given a point P, does there exist a point Q such that the line segment PQ has length equal to the circumference of a circle of radius of length 1?

Hurkyl

May11-04, 11:45 PM

Oh, I almost forgot to ask this again: how does your decimal system represent irrational numbers, like √2 or π?

Hurkyl

May11-04, 11:57 PM

Blarg, you got me with the red herring. :frown:

You've presented no reason whatsoever for one to suppose that your number system has anything to do with ordinary Euclidean geometry. (which is part of the reason why I was asking these particular questions!)

Thus, I am still interested in seeing you prove that your decimal system contains a number whose square is 2. (Since the proof from Euclidean geometry does not suffice)

Also, I really should rephrase my question about circle continuity:

In your system, I'm defining the "ram plane" to be ordered pairs of "ram decimals"; so each point has an x and y coordinate that is one of your decimal numbers.

I define the square distance between the two points (x, y) and (u, v) to be (x - u)^2 + (y - v)^2

So then, a "ram circle" would be the defined as I did above: Given a point O and another point R in the ram plane, for any point X I say:

Let x be the square distance between O and X.
Let r be the square distance between O and R.
Then, if x < r, X is said to be inside the circle OR
Then, if x = r, X is said to be on the circle OR
Then, if x > r, X is said to be outside the circle OR

ram2048

May12-04, 12:15 AM

i didn't MEAN to red herring you, as all it did was make more problems for me ;D

too much at once. just pick ONE

and my notational system for decimals only represents irrational decimals created from fractional values to create rational ones.

i could possibly devise a way to represent pi as an accurate decimal as well, but i don't intend to bite more than i can chew.

if by some miracle people accept THIS then we can work on new stuff

ram2048

May12-04, 12:15 AM

by the way, what's the accepted calculation for obtaining digits of pi ?

Zurtex

May12-04, 02:56 AM

by the way, what's the accepted calculation for obtaining digits of pi ?
Well ram2048 you have totally failed to answer most of the questions and any of the questions with answers that make sense because most of them do show you clearly wrong. There are many ways of calculating Pi, here is one:

\pi = \sum_{k=0}^{\infty} \frac{{-1}^k}{2k + 1}

matt grime

May12-04, 03:22 AM

i know nothing of continuums.

not going to even go there.

me: <looks up definition of continuum> if a continuum is blah blah then you could blah blah and so blah blah
you: no you can't
zurtex: WRONG hahaha you are WRONG SEE? YOU SUCK HAHAHA
me: <sad>

how about no?

so, you're going to spout on at length about what 'our' mathematical system is, and isn't, yet you aren't going to learn anything about it that might help in your understanding?

ram2048

May12-04, 03:54 AM

don't start, Grime

i'm taking this a step at a time, i don't need to be stretched in directions i don't have to unnecessarily.

and like i've said before, with my current knowledge i have already spotted flaws in the system and come up with solutions for them that work. i'll get to continuums when time and resources permit.

Well ram2048 you have totally failed to answer most of the questions and any of the questions with answers that make sense because most of them do show you clearly wrong.

hmm i am clearly wrong when i can do calculations that come up with the EXACT correct answer and you are RIGHT when you come up with approximations.

yes that makes ALOT of sense Zurtex.

Zurtex

May12-04, 04:10 AM

hmm i am clearly wrong when i can do calculations that come up with the EXACT correct answer and you are RIGHT when you come up with approximations.

yes that makes ALOT of sense Zurtex.
:rolleyes: You do realise "alot" isn't a real word?

As you say sums to infinity are wrong please give us how you would define \pi or \sqrt{2}. Also please define this so called default infinity. Also please could you show us if multiplying 10 by 0.\overline{9} "creates" a 0, which if you'd listened you'd know was absurd, where this 0 actually happens considering all the 9s go on forever.

These are just 3 things you can't do, your system isn't more accurate it is just plain wrong. I have tried to show you this as have many people, but you apparently haven't taken in a single point and just carry on spewing out rubbish.

Zurtex

May12-04, 04:28 AM

So you see all infinities are different and there is some infinity which follows this:

\frac{1}{\infty} = 0.\overline{0}1

As all numbers of the form 0.000...0001 can be written of the form:

\frac{1}{10^n}

Where n is a natural number. Such that n is how many digits the significant number 1 is below the unit digit. So taking your previous unique infinity can you define for us:

\log_{10} \infty

So we know how far down these infinitesimal numbers are. Once again if you think about this you will realise how absurd this whole thing is.

matt grime

May12-04, 07:13 AM

yes, ram, you've managed to sort out all the problems of the real number system without bothering to find out what the real numbers are, that's exactly what you've done. have you looked up any of these other things you've been told about where people have actually done these things properly? robinsonian analysis? infinitessimals, conways arithmetic of infinite cardinals?

for instance, associate a number n with a set of cardinality n, then addtion of n+m is the cardinality of the union of the sets, thus if we take infinity as aleph-0, for the sake of argument, then aleph-0+aleph-0= aleph-0=aleph-0 + 1.

of course w and w+1 are distinct in the ordinals (w is the first infinite ordinal).

but we wouldn't expect you to have checked that.

so go on, infinity is a 'real number' is it? which one?

Zurtex

May12-04, 07:25 AM

If multiplying by 10 "creates" a 0 then this will surely apply so that it "creates" a 0 in 10\pi. Also multiplying your default infinity by \pi will give us a number such that it will only have 0s as digits below the unit digit "after the decimal point". So please tell us in the multiplication:

[tex]\infty \pi[/itex]

What is the number in the unit digit \pi? Or another way of putting it is "what is the last digit" in \pi?

hello3719

May12-04, 07:25 AM

with my current knowledge i have already spotted flaws in the system and come up with solutions for them that work.

Can you show us and explain us these "flaws" in our system?
Show me 1 contradiction in our system and not something you don't like because you think it is illogical without even backing it out.

ram2048

May12-04, 12:05 PM

okay this is just stupid.

you: show me where we're wrong
me: <here> see?
you: you've proven nothing show me where it's wrong
me: wtf <here>
you: you know nothing, show me where we're wrong

i'm not showing you any more until you can come up with a disproof for my theory that consists of more than "that's not how we do things, you must be wrong" or "there is no end, you must be wrong"

my numbers are the EXACT same ones in subtraction as the ones gotten from multiplication. yours are NOT "blah blah 8.999~ IS 9..." if it was 9 then it would have calculated out to BE 9.

Zurtex

May12-04, 01:04 PM

ram2048 that is total rubbish. We have shown you over and over again why it is what it is and shown you over and over again why your so called disproofs are wrong. But you just keep posting over the same thing over and over again.

ram2048

May12-04, 01:19 PM

that consists of more than "that's not how we do things, you must be wrong" or "there is no end, you must be wrong"

no, not really

Zurtex

May12-04, 01:36 PM

no, not really
Then at least reply to the last few posts once again showing how unbelievably wrong you.

Edit: Why have you quoted yourself AGAIN? And disagreed with yourself.

ram2048

May12-04, 03:20 PM

Then at least reply to the last few posts once again showing how unbelievably wrong you.

translation: dance for us while we give you theoretical problems way over your head even though they have nothing to do with the issue at hand.

prove me wrong first.

i wasn't quoting myself i was re-iterating my current stance and the disagreement was with YOUR statement as a response.

if you would at least read and understand what i'm telling you, you wouldn't ask such dumb questions :D

matt grime

May12-04, 04:08 PM

Ok, here's a proof that 0.99999....=1

0.9999... is clearly the limit of the sequence x_n=0.9..9 with n nines in it, do we agree on that, since that isby definition what the symbol 0.9999..... means, there can be no argument there.

consider the sequnce y_n=1 for all n.

y_n must tend to 1, yes?

consider x_n-y_n, then in abs values this is less than 1/10^n, hence x_n-y_n converges to zero, agreed? thus by the definition of the real numbers, these sequences lie in the same equivalence class of cauchy sequences, hence they are equal as real numbers.

now, point by point go through that and state where it is wrong, not where you think it is wrong, but where you can prove it is mathematically wrong.

ram2048

May12-04, 04:24 PM

0.9999... is clearly the limit of the sequence x_n=0.9..9 with n nines in it, do we agree on that, since that isby definition what the symbol 0.9999..... means, there can be no argument there.

.999~ has a limit of 1 but a SUM of .999~

it TENDS to 1, but it will never have a value of exactly 1.

.333~ has a limit of 1/3 but a SUM of .333~

it tends to 1/3 but will never have a value of exactly 1/3

matt grime

May12-04, 04:29 PM

Hurkyl

May12-04, 04:29 PM

too much at once. just pick ONE

Asking ram decimals to represent the square root of 2 is probably going to be the most challenging "simple" question we could ask.

One problem is that you need to be able to forbid certain decimals from being in your system (such as the standard meaning to 0.999~ or 0.333~). However, you need to retain enough flexibility to represent irrational numbers like sqrt(2).

However, the most disasterous part, I think, is that without the nicely repeating patterns of rational numbers to help you out, there may be no way to "connect" the end of a decimal to the beginning of a decimal.

Note that your system already has limited forms of this problem with rational numbers; for instance, for the decimal 0.121212~, is the last digit a 2 (with remainder 12/99) or is it a 1 (with remainder 21/99)? This problem is solvable; I think the irrational version is not (at least not without using some difficult mathematics).

by the way, what's the accepted calculation for obtaining digits of pi ?

There can be particular methods for particular numbers, but there is a most general method to constructing a (mathematical) decimal, which I will demonstrate to construct the first few digits of sqrt(6).

2 < sqrt(6) < 3, so the first digit is 2
2.4 < sqrt(6) < 2.5, so the second digit is 4
2.44 < sqrt(6) < 2.45, so the third digit is 4
2.449 < sqrt(6), so the fourth digit is 9
2.4494 < sqrt(6) < 2.4495, so the fifth digit is 4
...

(see 1: below)

This recursively defines the n-th digit (for any positive integer n). You'll have to devise some clever trick, though, to extend this definition to go beyond integer positions, and such extensions are not generally easy, especially with recursions that strongly depend on knowing the previous term.

"blah blah 8.999~ IS 9..." if it was 9 then it would have calculated out to BE 9.

Do you also assert that the fraction 2/4 cannot be equal to the fraction 1/2?
(After all, if it was 1/2, then it would've been calculated out to BE 1/2...)

1: For simplicity, I was not considering the possibility that any of these partial decimals may be equal to the target number, but you wouldn't be happy with such a more complete method anyways, because it would yield yet another direct proof that 0.999~ = 1

ram2048

May12-04, 04:36 PM

then you don't understand analysis do you? And we can see where you've gone wrong. No finite partail sum will reach the limit, but so what? this is your issue with a DEFINITION, sorry, you don't like it, but the point of analysis is to allow one to take limits.

you've not disproved anything, just shown you don't accept a definition. we can't alter that, but that's your problem

see the problem lies within that limit of infinity.

you: the limit is 1, therefore when infinity is hit, it equals 1.
me: the limit is 1, but even when infinity is hit it doesn't equal 1.

logically i am correct. but you win out with practicality because there's no good reason to go "beyond" infinity.

you define your lim as the sum of the set to infinity, but there's no point in you saying that because you've already limited yourself mentally at infinity.

hello3719

May12-04, 04:41 PM

so still think that infinity +1 > infinity,
how can you go beyond infinity if infinity is not a number?

ram2048

May12-04, 04:42 PM

Do you also assert that the fraction 2/4 cannot be equal to the fraction 1/2?
(After all, if it was 1/2, then it would've been calculated out to BE 1/2...)

no because 2/4 can be simplified to a rational base expression. if you could simplify ([8/1].[9/1][9/1][9/1]~[1/1]) any of those terms then they should be simplified... :D

i'm just messin with ya, i'm working on sqrt2 at the moment

Hurkyl

May12-04, 04:44 PM

you: the limit is 1, therefore when infinity is hit, it equals 1.

When did we say "when infinity is 'hit' " or anything similar?

As long as you are in the mindset that our definitions are literally involving the behavior at infinity, you will never understand them.

matt grime

May12-04, 04:47 PM

we haven't limited ourselves at all to stopping at infinity, which in the sense you mean is not what we're doing anyway as we never reach infinity, what with it not being an integer and everything. if you'd read anything about transfinite induction as i'd suggested rather than dismissing it contemptuously you'd realize

ram2048

May12-04, 04:57 PM

well that makes no sense.

.999~ tends to 1. but since 1 can never be reached it IS 1?

instead of

.999~ tends to 1. but since 1 can never be reached it never becomes 1.

you seriously believe that?

and don't say you're not limiting yourself at infinity when you clearly are.

.999~ having infinite number of digits 9. if i add one more at the end i come EVEN CLOSER to 1. but you won't allow that so hence LIMIT.

matt grime

May12-04, 05:10 PM

if you want to add another nine after the infinite number of them already there then you aren't using decimal notation so you aren't making any sense.

every mathematical object is limited by its definitions, but that you interpret that lmit in a negative way is your own problem: there is no square root of -1 in the real numbers, does that imply they are analytically deficient?

Hurkyl

May12-04, 05:12 PM

ram2048

May12-04, 06:09 PM

.999~ = 1 because:

For every positive real number e,
there exists a positive integer N such that,
for every integer m bigger than N,
|1 - 0.9..9| < e (there are exactly m 9's in that number)

Notice, in particular, that this definition is entirely in terms of finite examples. Infinity doesn't inter into this at all.

i have no idea where you're trying to go with that one, i'm putting positive reals in for e but it doesn't seem to be telling me anything useful :(

hello3719

May12-04, 06:33 PM

1/9 = 1/10 + 1/100 + 1/1000 + ...

then 9 * 1/9 = 9/10 + 9/100 + 9/1000 +...
9/10 + 9/100 + 9/1000 +... = .999...
9/9 = .999...
1 = .999...

tell me where do you see a contradiction in this proof.( show me the illegal step )

ram2048

May12-04, 06:43 PM

1/9 = 1/10 + 1/100 + 1/1000 + ...

right there. convergent sum to infinity.

.111~ has a limit of 1/9 but a sum of .111~ [edit] forgot the ~ thingy

in my notation 1/9 = .111~ r1/9
9 x .111~ r1/9 = .999 r9/9 = 1
and
9x 1/9 = 1

perfectly.

yours

9x 1/9 = 1

BUT

9x.111~ = .999~

Hurkyl

May12-04, 06:44 PM

Were you able to find an appropriate N for every positive real you plugged in for e?

What if we asserted 0.999~ was something other than one... like, maybe, 1.01, which (by definition!) would mean:

For every positive real number e,
there exists a positive integer N such that,
for every integer m bigger than N,
|1.01 - 0.9..9| < e (there are exactly m 9's in that number)

I bet you can find an e that is a counterexample!

ram2048

May12-04, 06:50 PM

where the heck is n in that equation.

damn don't play hide the variables with me :|

hello3719

May12-04, 06:55 PM

1/9 = 1/10 + 1/100 + 1/1000 + ...

right there. convergent sum to infinity.

.111~ has a limit of 1/9 but a sum of .111

in my notation 1/9 = .111~ r1/9

Remember we are talking about OUR system not your system, you are using your definitions here to contradict the proof, don't introduce your notations in our system. Use our definitions of infinity and real numbers to find a contradiction. If you can't then this is the proof is completely valid and 1= .999... Well before asking this, do you know how do we define infinity and do we treat it ?

ram2048

May12-04, 07:02 PM

hello3719

May12-04, 07:15 PM

i'm using your system and it doesn't work that's why i introduced my system :P

9x.111~ in your system = .999~

9x1/9~ in your system = 1

this leads to TWO possible conclusions.

.999 is equal to 1

OR

1/9 ≠.111~

you have chosen the 1st possibility and not proven it, i have chosen the second possibility and HAVE proven it.

dunno what more i can tell you...

k this means that you didn't understand the proof and the properties of our system integrated into it. 1/9 = .111... isn't a conlusion you can take from the proof but a CONSEQUENCE OF OUR SYSTEM if you didn't understand that yet.
You really don't know how to analyse a proof. 1/9 is equal to .111... in our system by using the definition of our infinity in the concept.

Hurkyl

May12-04, 07:21 PM

Let's go over the (usual) definition of the limit of a sequence again.

\lim_{n \rightarrow \infty} a_n = L

means:

For every positive real number e
there exists a positive integer N
such that for every integer m bigger than N,
|L - a_m| < e.

(When I say "means", I really mean it; I don't mean "is assumed to be", "is proven to be", or "is approximately")

So, for example, if I claim:

\lim_{n \rightarrow \infty} 1/n = 0

That means that I claim

For every positive real number e
there exists a positive integer N
such that for every integer m bigger than N,
|0 - 1/m| < e

A proof of this would go as follows:
Let e be any positive real number.
Select N to be the first integer bigger than 1/e.
Let m be any integer bigger than N.

Then, we have:
m > N
N > 1/e
m > 1/e
1/m < e
|-1/m| < e
|0 - 1/m| < e

So we have proven:
for any positive real number e,
there exists a positive integer N such that
for any integer m bigger than N,
|0 - 1/m| < e

Thus, it is true that

\lim_{n \rightarrow \infty} 1/n = 0

Because this means precisely what we have proven.

ram2048

May12-04, 07:46 PM

can you please run some numbers through that for me? still can't find the n. is it in another equation or is n the same thing as m?

Hurkyl

May12-04, 08:04 PM

Maybe the "game" intepretation will help you.

Suppose we play this game:
You pick a positive real number (call it e).
I pick a positive integer (call it N).
You pick a positive integer bigger than N (call it m).
If |L - a_m| < e I win, otherwise you win.

If I can win all the time, then it is true that:

\lim_{n \rightarrow \infty} a_n = L

If you can win all the time, then it is false.

So, for example, let's consider the case where L = 0 and a_n = 1/n. Then, the "game" is:

You pick a positive real number (call it e).
I pick a positive integer (call it N).
You pick a positive integer bigger than N (call it m).
If |0 - 1/m| < e I win, otherwise you win.

If I win every time we play then
\lim_{n \rightarrow \infty} 1/n = 0
is true.

(And I will win every time. :smile: My strategy is to pick N to be the first integer bigger than 1/e)

Zurtex

May12-04, 08:57 PM

translation: dance for us while we give you theoretical problems way over your head even though they have nothing to do with the issue at hand.

prove me wrong first.
Done it, many times. Go back to the first page and you shall see.

matt grime

May13-04, 05:26 AM

The bit where you don't understand what the definition of the real numbers is is the bit that I meant you should perhaps take in from all this.

In the proof i offered that 0.99...=1 in the real numbers used the definition of what the rel numbers are - they are not decimal expansions, they are equivalence classes of cauchy sequences of real numbers. (or dedekind cuts, but that's less usefulo here, but they are equivalent.) When you've learnt all the meaning of all those terms perhaps then you ought to reconsider your position. and seeing as you don't even follow the idea of a limit we culd be here some time.

x_n=0.9...9 n nines
y_n=1 for all n

are different cauchy sequences, but they have the same limit, and thus by the defintion of real numbers represent the sam real number. if not then you have the problem in your system that 1/n does not tend to zero (since you insist that these infinitesimals are real numbers), indeed you can't do analysis at all because practically nothing converges or has a decimal expansion, moreover not even multiplication is continuous., and subsequences of convergent sequence tend to different limits then the sequence's limit.

ram2048

May13-04, 05:42 AM

and what you don't seem to understand is that the number .999~ is on the same slate as infinity and thus is subject to your same limitations.

1. there is no "Archemedian" number closer to 1 than .999~ because 9 is the digit closest to 10 and you are not allowing the addition of more 9's on the end.

2. Archemedian properties wouldn't apply to numbers subject to a limit in the first place. for example let's limit 500 to be the highest possible number. what's the closest number to 0 that is not 0. 1/500 is definately NOT zero, but you cannot make anything closer. By archemedian then 1/500 is EQUAL to 0. which we KNOW is false.

3. .999~ is not the closest number to 1 in base 11. i don't have a calculator handy. The number closest would be .aaa~ where a is the 11th digit in that base.

4. subsequences of my set do not converge to different limits, they converge to the same one, but at a different rate/staggered interval if started at the same "time"

Hurkyl

May13-04, 06:36 AM

1. there is no "Archemedian" number closer to 1 than .999~ because 9 is the digit closest to 10 and you are not allowing the addition of more 9's on the end.

(I don't know what an Archmedian number is; I'm guessing that's your new name for the real numbers because they have the Archmedian property)

No... we don't allow the addition of more 9's because EVERY position (to the right of the decimal place) already has a 9; there's no digit that we could change into a 9.

Archemedian properties wouldn't apply to numbers subject to a limit in the first place. for example let's limit 500 to be the highest possible number.

I'm not even sure it makes sense to talk about the Archmedian property applying to this system, because it fails to have many basic properties of number systems. (e.g. addition of two numbers is frequently undefined)

But, indeed, I can say that this system has no integer which is larger than 500, so one might be inclined to say that it fails to have the Archmedian property.

what's the closest number to 0 that is not 0. 1/500 is definately NOT zero, but you cannot make anything closer. By archemedian then 1/500 is EQUAL to 0. which we KNOW is false.

No... remember that the proof that (in the real numbers, or any ordered field) there is no smallest number larger than zero doesn't have anything to do with the Archmedian property; it goes like this:

Assume there is a smallest number larger than zero. Call it x. But then 0 < (1/2) x < x, which contradicts our assumption. Thus, there cannot exist a smallest number larger than zero.

This proof fails in the system you describe because we cannot multiply 1/500 by (1/2).

3. .999~ is not the closest number to 1 in base 11. i don't have a calculator handy. The number closest would be .aaa~ where a is the 11th digit in that base.

Correct. In base 11, .999~ [11] = 9/10 [11] = 9/11 [10] = .8181~ [10]. (numbers in brackets indicate the base in which the number is written)

And, incidentally, .aaa~ [11] = 1 [11]

subsequences of my set do not converge to different limits, they converge to the same one

This would suggest that there is only one positive infinite value...

If we take two sequences, a(n) and b(n), each of which converge to some positive infinite value, then we can make a new sequence c(n) by:
c(2n) = a(n)
c(2n+1) = b(n)
If c also converges to some positive infinite value, then, because a and b are subsequences of c, a and b must have the same limit.

I guess you can avoid this proof that there is only one positive infinite value (as far as ram limits are concerned) by forbidding some (most, actually) sequences that grow without bound from converging to infinite numbers.

matt grime

May13-04, 07:08 AM

seeing as subsequences converge to the same ram number, we can demonstrate another way that there is no smallest non-zero positive ram number:

1/n converges to something, call i x, so does 1/n^2, and is must also converge to x^2, thus x=x^2, and x=0 or 1, so it must be 1. thus, given any non-zero ram number there is a number of the form 1/n less than it. oops.

of course this is assuming you are claiming your numbers are an improvement on the reals. and if not then what's you're point? have you read about robinsonian analysis yet? you might learn something if you weren't closed minded about maths like you are (we aren't by the way because we qualify all of our statements properly, and fully admit their inherent limitations, which is what generalizations are for).

why do you still keep pretending we think infinity is the largest real number? we've explained it isn't, if yo'ure are going to deliberately misrepresent us you'll always get it wrong

ram2048

May13-04, 07:55 PM

No... remember that the proof that (in the real numbers, or any ordered field) there is no smallest number larger than zero doesn't have anything to do with the Archmedian property; it goes like this:

Assume there is a smallest number larger than zero. Call it x. But then 0 < (1/2) x < x, which contradicts our assumption. Thus, there cannot exist a smallest number larger than zero.

This proof fails in the system you describe because we cannot multiply 1/500 by (1/2).

0 < (1/2 x 1/∞) < 1/∞

same problem. same limit. which is why i was using that as an example.

you guys use limits all the time yet you don't understand one when you see it?

BECAUSE infinity is your limit, you cannot get a number closer to 1 than .999~ but that does not mean they're equal because by imposing that limit you also cancel out the validity of using Achemedian Property to assert that equality. With NO limit on infinity it's very easy to see that Infinity+1 digits of 9 is closer to 1 and infinity+2 digits is even closer... ad infinitum 1 can never be reached and you can archemedes all you want...

Hurkyl

May13-04, 11:11 PM

0 < (1/2 x 1/∞) < 1/∞

same problem. same limit. which is why i was using that as an example.

you guys use limits all the time yet you don't understand one when you see it?

BECAUSE infinity is your limit, you cannot get a number closer to 1 than .999~ but that does not mean they're equal because by imposing that limit you also cancel out the validity of using Achemedian Property to assert that equality. With NO limit on infinity it's very easy to see that Infinity+1 digits of 9 is closer to 1 and infinity+2 digits is even closer... ad infinitum 1 can never be reached and you can archemedes all you want...

How is it the same problem? I can identify some crucial differences that prevent the analogy from holding:

(a) In your toy example, the "limits" 500 and 1/500 are both part of the number system. ∞ and 1/∞, however, are not part of the real numbers.

(b) In your toy example, operations like + and * are not defined for every pair of numbers. However, + and * are defined for every pair of real numbers.

(c) In your toy example, the limit was "reachable"; e.g. you could do arithmetic that exceeds the limit (if it were allowed). No such problem exists in the real numbers. (This is merely a conceptual restatement of (b). )

Furthermore

(d) You are confusing the word "limit" (as in a limit is something that cannot be surpassed) with the word "limit" (as in a limit is something to which a sequence converges)

(e) Did you not notice that the proof I supplied does not use the Archmedian property? The proof works for all ordered fields, including those that are non-Archmedian.

(f) Things are only "easy to see" because:

(f1) You insist on treating "infinity" as if it behaves identically to finite things.
(f2) You haven't supplied a definition to which you intend to adhere, thus you aren't restricted by petty things like logical consistency.

matt grime

May14-04, 03:35 AM

WHen i see 0.999... i know what that means. i add 9/10 to 9/100 to 9/1000 etc, now when you have your thing that's 'closer' to 1, which we'll notate as 0.9....|9 for the sake of a better argument, i think i need to add 9/10,9/100 and so on but what does the 9 after the bar mean i should add? 9 over 10^what?

ram2048

May14-04, 05:24 AM

matt grime

May14-04, 05:33 AM

ram, you miss the point entirely. i know what decimal notation means, and i know how to use it.

now you want in your system to have 0.999999.... an infinite number of nines, and then to add one more on to the right. so what does that mean? what are you adding in terms of numbers? the 9 at the n'th spot in the proper expansion means add 9/10^n, so what is it you think you are adding on in your notation when you add one more 9 to the right of all the others. you must be able to tell us.

matt grime

May14-04, 05:38 AM

you also don't appear to understand the idea of cardinality. when you say the number of nines is infinity plus one, you are using cardinals, and implictly we are saying there are aleph-0 of them, and in the arithmetic of cardinals aleph-0 + 1 = aleph-0, it is the cardinality of the set of naturals union a singleton set which is the same as the card of the set of naturals.

saying 0.9999....=1 follows from the definition of the real numbers and decimal expansion, it is not a convenience.

note also that 3 is not 6, yet mod 3 they are equivalent....

matt grime

May14-04, 05:48 AM

if you COULD get to infinity

as we can't do that in the real numbers your argument is vacuous

hello3719

May14-04, 05:44 PM

Ram2048, k then how would you represent

1/3 + (1/3)^2 + (1/3)^3 + ... + (1/3)^n as n goes to infinity

do you agree that it converges to one specific number ? if yes, then try to write that number.

pallidin

May14-04, 06:21 PM

Ram, how can it be made more clear?? Infinity has no specific numerical value and it never can have one. Infinity is a mathematical concept of relationship that is dynamic; it changes and changes and changes... always increasing in numerical value for positive infinity, or always decreasing for negative infinity.
There is no end to this sequence, because it is exactly that: A concept based entirely on having no end.

pallidin

May14-04, 06:48 PM

pallidin

May14-04, 07:02 PM

We are go for launch: 3...2...1...0... lift-off.

Yeah, mathematically incorrect, as it would suggest: 3...2..1... wait for infinity... lift-off.

So, instead they say. 3...2...1... lift-off.

Get it?

ram2048

May14-04, 07:05 PM

To the paradox of always being able to divide a distance in half, thereby never mathematically reaching it's destination(for a collapse event):
In standard mathematical division this paradox is correct, and no supercomputer on earth could EVER collapse a halving division sequence into zero, or "contact"
But where is the flaw?
Halving to zero is inherently impossible because zero is, itself, an infinity. So, "contact" cannot occur at zero, because at zero THERE IS NOTHING TO CONTACT WITH!!
So, contact must occur at a defined numerical value specific to the event, and zero is NOT a numerical value, it is an infinity. As such, we can arbitrarily define the "contact" value as "greater than zero", thus, collapse is possible.

Palladin: but that would work the same way for convergence to ANY value, not necessarily with zero, so that rationalization holds no significance.

9/10 +9/100 +9/1000 converges to 1 but can NEVER reach 1. these guys will continue to try and confound you with mathematical notations all which have no basis in logic whatsoever, but the truth of the matter is simply that 9 is NOT 10 and infinite 9's is not 1.

but that issue is concluded, working on root 2 now :D

forgot what the heck i'm supposed to do with the thing, however....

ram2048

May14-04, 07:14 PM

Matt: cardinality doesn't enter into it. by defining a default infinity i'm creating a new numberline to extrapolate logical numerical transforms on.

i can say infinity(d) is my 0 and everything i do to it is taken from that point.

certain things will logically transform back to the realm of known reals, like infinity(d) - infinity(d) = 0 or infinity(d)/infinity(d) = 1.

pallidin

May14-04, 07:42 PM

Palladin: but that would work the same way for convergence to ANY value, not necessarily with zero, so that rationalization holds no significance.

9/10 +9/100 +9/1000 converges to 1 but can NEVER reach 1. these guys will continue to try and confound you with mathematical notations all which have no basis in logic whatsoever, but the truth of the matter is simply that 9 is NOT 10 and infinite 9's is not 1.

but that issue is concluded, working on root 2 now :D

forgot what the heck i'm supposed to do with the thing, however....

I am fully aware that it works for both convergence and divergence. I was simply stating an example on one aspect to keep it short.

Ram, perhaps you are not understanding certain things correctly, and I do mean that gently.
Basically, you are suggesting an infinite mathematical series within a finite mathematical framework. This is not possible.
You are certainly correct that .999 forever cannot equal 1. OK. So what? 1.999~ will never equal 2. Agreed, and no one has a problem with that.
But, do you see what you are doing?
Your are forcefully introducing an infinity(.999 ~) in a direct relationship with a finite value(1) on your same number sequence scheme.
This cannot be done, because you are eventually saying that infinity and finity co-exist. Even if it could, under co-existance, what happens?
Think about it.
Infinity takes over by virtue of property and finity becomes non-existant.

Hurkyl

May14-04, 07:53 PM

... There is no end to this sequence, because it is exactly that: A concept based entirely on having no end.

Mathematicians are somewhat less vague about how they define things. For the record, the symbols +∞ and -∞ are defined precisely so that they are the endpoints of the extended real line. Topologically, extending the real numbers to include +∞ and -∞ is exactly analogous to extending the interval (0, 1) to include 0 and 1.

but the truth of the matter is simply that 9 is NOT 10 and infinite 9's is not 1.

2 is not 1, and 4 is not 2, and 1/2 is not 2/4, yet these manage to be equal...

Hurkyl

May14-04, 08:31 PM

Ok, a reminder for both ram and pallidin:

Mathematically, a decimal is a function from the integers into a set of digits. If f is a decimal, we interpret f(n), intuitively, to mean "the digit in the n-th place of f".

In particular, 0.999~ is the function f where:
f(n) = 0 (for n >= 0)
f(n) = 9 (for n < 0)

In this way, we've managed to represent an "infinite mathematical series within a finite mathematical framework".

Arithmetic can be defined on these decimals representations directly, rather than defining it by translating decimals into real numbers, doing the arithmetic, then translating back to decimals. For example, addition would then be defined as (and this is just arithmetic you learned in elementary school, except it computes from left to right):

Suppose f and g are decimal numbers. We define a new decimal, h, to be the sum of f and g by:

For each position m:

(Computing the carry)
If it is never true that f(n) + g(n) >= 10 for n < m, then carry = 0.
Otherwise, choose n to be the highest integer smaller than m
so that f(n) + g(n) >= 10. If it is true that f(p) + g(p) = 9 for all
p between m and n, then carry = 1. Otherwise carry = 0.

(Computing the digit)
If f(m) + g(m) + carry < 10, then we set h(m) = f(m) + g(m) + carry.
Otherwise, we set h(m) = f(m) + g(m) + carry - 10.

For example, let's add .451 and 1.573
First, let's recall that these decimals are really functions from the integers into digits, so let's write them down. Let's call the first f and the second g:
f(n) = 0 (n >= 0)
f(-1) = 4
f(-2) = 5
f(-3) = 1
f(n) = 0 (n <= -4)

g(n) = 0 (n >= 1)
g(0) = 1
g(-1) = 5
g(-2) = 7
g(-3) = 3
g(n) = 0 (n <= -4)

So now, let's compute h := f + g.

Notice that n = -2 is the only place where f(n) + g(n) >= 10, and
n = -1 is the only place where f(n) + g(n) = 9.

Let's first compute h(m) for m >= 1.
The carry is zero because 0 is between m and -2, and f(0) + g(0) < 9.
Furthermore, f(m) + g(m) = 0, therefore h(m) := 0.

When computing h(0), the carry is 1 because f(-2) + g(-2) >= 10, and
f(n) + g(n) = 9 for all n between 0 and -2. Therefore, h(0) = f(0) + g(0) + 1 = 0 + 1 + 1 = 2.

Similarly, the carry is 1 when computing h(-1). (There is no n between -1 and -2, thus it is vacuously true that f(n) + g(n) = 9 for all n between -1 and -2)
Furthermore, f(-1) + g(-1) + carry >= 10, so:
h(-1) = f(-1) + g(-1) + carry - 10 = 4 + 5 + 1 - 10 = 0

And then we get h(-2) = 2, h(-3) = 4, and h(n) = 0 for n <= -4. Thus,
we've proven that

.451 + 1.573 = 2.024

Now, let's do another example; 0.999~ + 0.111~.
Again, let's write these as functions:
f(n) = 0 (n >= 0)
f(n) = 9 (n < 0)
g(n) = 0 (n >= 0)
g(n) = 1 (n < 0)

When we apply the addition algorithm to compute the sum, h, we get:
h(n) = 0 (n >= 1)
h(n) = 1 (n <= 0)

Note, however, that when we apply the addition algorithm to 1 + 0.111~ we get:

h(n) = 0 (n >= 1)
h(n) = 1 (n <= 0)

Since we get the same function in both cases, we get the same decimal number. 0.999~ + 0.111~ = 1 + 0.111~

This is the "use" of defining that 0.999~ = 1. If we did not define it to be so, then addition as I've defined it above doesn't behave like addition is "supposed to behave". However, if we make this identification, one can then prove that all of the arithmetic algorithms we learned in elementary school behave how they're supposed to behave (e.g. that dividing by 9 then multiplying by 9 is a no-op).

One thing to note is that the elementary school division algorithm involves computing a remainder once you stop; having a remainder turns out to be unnecessary with the decimals in the standard mathematical definition. That they are unnecessary relies entirely on putting the pseudoparadoxes of nonfininte cardinals and ordinals to practical use.

ram2048

May14-04, 08:54 PM

Now, let's do another example; 0.999~ + 0.111~.
Again, let's write these as functions:
f(n) = 0 (n >= 0)
f(n) = 9 (n < 0)
g(n) = 0 (n >= 0)
g(n) = 1 (n < 0)

When we apply the addition algorithm to compute the sum, h, we get:
h(n) = 0 (n >= 1)
h(n) = 1 (n <= 0)

Note, however, that when we apply the addition algorithm to 1 + 0.111~ we get:

h(n) = 0 (n >= 1)
h(n) = 1 (n <= 0)

Since we get the same function in both cases, we get the same decimal number. 0.999~ + 0.111~ = 1 + 0.111~

assuming equal infinity number of digits <for simplicity "infinity(d)"> of each non-terminating decimal. (which would be the logical and sane way to compute that sum).

.999~ + .111~ =

h(n) = 0 (n >= 1)
h(n) = 1 (n <= 0 BUT > h(n)=-Infinity(d) )

that computation yields 10, not 11.

granted it's off the chart of any computer, but there's still no denying that 1 + 9 = 10.

Hurkyl

May14-04, 10:16 PM

assuming equal infinity number of digits

I presume you are referring to the nonzero digits.

The set of positions where 0.999~ has a nonzero digit is the same as the set of positions where 0.111~ has a nonzero digit (in particular, that set is exactly the set of negative integers). What do you think needs to be "assumed"?

<for simplicity "infinity(d)">

How is that simplicity? You've:

yet to fully define that term (either via a list of axioms or by fully specifying just what it is)

you've given no reason to think that it has anything to do with "numbers of things"

if it does have to do with "numbers of things", you've given no reason to think that it should correspond to the number of negative integers

(which would be the logical and sane way to compute that sum).

The logical way to compute the sum would be either directly from the definition of addition, or some alternative algorithm that is equivalent to the definition. I happen to think the definition is a sane method as well, because it corresponds directly to the way we learned how to add decimals in elementary school.

(n <= 0 BUT > h(n)=-Infinity(d) )

First, I presume you meant "n <= 0 but n > -Infinity(d)"

Secondly, I'm not sure about the point you are making by bringing this up. All integers are greater than -Infinity(d), and since n is restricted to the integers, it is fairly pointless to specify that n > -Infinity(d), since it is true for all n.

that computation yields 10, not 11.

What computation? Are you forgetting about carry?

ram2048

May15-04, 12:55 AM

the very first one to the far right. 9 + 1 with no carry

in all arithmetic answers stem from the right, that's how we do things.

111
+999
1110 <-- note the 0

your assumption that a 1 would carry for that digit has no basis in logic because there will always be a digit for which there is no carry.

this is assuming exactly infinite(d) number of decimal digits for both numbers added. (logical)

and yes i have "defined" my default infinity. twice in this thread i believe. and have proven that it DOES work and yields perfect numbers as well.

i totally expect you to immediately come back and say "where" as you normally do so i will wait for that expectantly... :|

matt grime

May15-04, 01:48 AM

you may feel you have defined it but you haven't done so in any rigorous sense (listen to a mathematician for once). in particular you've not explained what number you get when you have infinity+1 9s in the expansion .9999....|9.

so to repeat myself again: i know what 0.999..... is, but what on earth does it mean for you to add one more 9 to the right of all the others?

as you say the number of 9s is infinite you are talking about cardinality. go and read your own post.

also please answer the other issue raised: clear 1 is not 2, 2 not 4, but 1/2 is 2/4. if you understood equivalence relations (of cauchy sequences or rationals! or in this case of localizations) you might begind to understand wherein your ignorance lies.

moreoever, just because you only know how to add from the right doesn't mean that we can't add from the left, it is a very easy algorithm to define, you might care to try it at some point

lvlastermind

May15-04, 02:21 AM

Matt. if .999~ = 1, BUT you said you believe in larger than infinite, then infinity+1 number of 9's = what? 1.000~9? infinity+2 9's = 1.000~99?

no matter how mant 9's you add you never get to 1. it doesn't even enter into it what 9 over 10^n the digit location is in. 9 is NOT 10.

and Hurkyl the differences you state have no relevance, what matters is there is a stated upwards limit on the number. And about not being able to GET to infinity but you can GET to 500, you should be able to draw a logical conclusion that if you COULD get to infinity, the result of 1/infinity would NOT be 0 because 1/500 or 1/100billion or 1/100bazomajillion with the same upwards limit would not be 0.

saying .999~ = 1 and 1/infinity=0 is a convenience. nothing more

---Proof that 2=1 if division by zero was allowed---

a = x [true for some a's and x's]
a+a = a+x [add a to both sides]
2a = a+x [a+a = 2a]
2a-2x = a+x-2x [subtract 2x from both sides]
2(a-x) = a+x-2x [2a-2x = 2(a-x)]
2(a-x) = a-x [x-2x = -x]
2 = 1 [divide both sides by a-x]

All of the steps are perfectly legal except for the last one, dividing both sides by a-x. What is a-x? Well, a=x (step 1), so a-x=0. In the last step, we divided by zero. That's not allowed. And this puzzle is a good example of why
it is not allowed.

---1/infinity not allowed---

The reason why 1/infinity is not allowed is because infinity doest have a value. Infinity can be .33333... or it can be .5555555... or it can be .888888... The point is that 1/.333333... and 1/.55555555.... and 1/.8888888... would yield different answers thus making it undefined.

---Infinity not real number---

Infinity is not a number, anyway (not in arithmetic or algebra, see Transfinite Numbers). We can add infinities, and multiply them (sort of). But, we don't get bigger infinities, we get the same infinity. That's interesting. But, if we multiply infinity to both sides of an equation, we are in big trouble. It is the same as dividing by zero. In our little puzzle, when we divided both sides by a-x, that was the same as multiplying both sides by infinity. It is meaningless. It is not allowed in mathematics.

Arithmetic with infinity is not allowed, because infinity is not a number. And, just like our little puzzle, we get answers that make no sense. Calculus is essentially the field in which we deal with infinity and division by zero. And, we never deal directly with infinity or division by zero. We always see what happens when a number gets large without bound or gets closer and closer to zero.

http://www.jimloy.com/algebra/two.htm

---Why 1/0 equals infinity---

What's 1/0? Infinity, right? We said above that we can't divide by zero. But, can't we divide by zero, if we're careful? Let's look at 1/0, more closely. In Calculus, we deal with problems like this by using limits. In other words, we don't look at 1/0, we look at 1/x (the graph of y = 1/x is shown) when x gets close to zero. Well, when x gets close to zero(asymptote), 1/x gets very large without bounds it is infinity. Not so fast, x also gets close to zero on the negative side. Then 1/x becomes a very large negative number, without bounds, it is negative infinity. So, the answer to the question, "What is 1/0?" is "plus-or-minus infinity." Kind of a wild answer, isn't it? It is not exactly simple.

---Why Division by infinity is not allowed...---

If division by infiniy were allowed then 2 would equal 1.

(1)infinity=(2)infinity
(divide both sides by infinity(if infinity were a real #))

1=2

---IN SIMPLE---

1/0 and 1/infinity are not allowed because you cannot divide an 1 apple into zero peices just the same as you cannot divide an apple into infinite peices.

ram2048

May15-04, 05:51 AM

matt you wouldn't even understand my cardinals if i told you. you'd be all "infinity(d+1)" isn't a number blah blah or "infinity(e)" isn't defined.

so what are you pushing for?

and |v|astermind: in my system (1)Infinity(d) = (2)Infinity(d) is false so stuff like that wouldn't happen.

default infinity is very functional like that.

matt grime

May15-04, 06:18 AM

for the fourth time: you say that you can get closer to 1 by adding another 9 after all the nines in .99999....

call this number 0.9...|9

but which real number does that represent?

i am perfectly capable of knowing that w and w+1 are distinct ordinals. you have not defined how to add two (infinite) cardinals, or what that addition represents. for instance in the usual arithmetic of the cardinals the sum of A and B is the cardinality of the union of sets of individual cardinality A and B. Thus aleph-0 +1 = aleph-0, yet w the first infinite ordinal is not equal to w+1.

Hurkyl

May15-04, 10:08 AM

i totally expect you to immediately come back and say "where" as you normally do so i will wait for that expectantly... :|

Yes, I am going to say "where", because that is your #1 problem. I have this wildly optimistic hope that one of these times I'll say it and you'll understand why it's a problem.

An integral part of why the decimal system works is because the positions are indexed by integers, and that there is no largest nor smallest integer.

And you are talking about digits at positions like infinity(d), but there can't be digits at those positions because they're not integers.

Zurtex

May15-04, 10:11 AM

O.K, just answer me these two questions please, if I multiply \pi by 10 will it create a "0"?

If so does that not mean there is a last digit to \pi?

ram2048

May15-04, 07:05 PM

it absolutely WILL make a "0"

we may never be able to compute to that digit EVER, but it does exist

ram2048

May15-04, 07:12 PM

And you are talking about digits at positions like infinity(d), but there can't be digits at those positions because they're not integers.

that's why it's a NEW SYSTEM. as i have stated MANY a time.

"your system won't work because it's not like ours"

well if it was exactly like yours what would be the point of claiming it was MY system... OR new?

:|

it works, and one day your math will be obselete.

Hurkyl

May15-04, 07:39 PM

You seem to have difficulty making clear when you are talking about the standard system and when you are talking about your system.

matt grime

May16-04, 03:35 AM

Zurtex

May16-04, 04:07 AM

it absolutely WILL make a "0"

we may never be able to compute to that digit EVER, but it does exist
Then you are saying there is a last digit to \pi :rofl:

This is what like the 60th time your argument has now been disproved?

ram2048

May16-04, 05:19 AM

as you claim to have a 'better' system for the real numbers, it ought to exactly concur with the real numbers (wherein 0.999...=1, so you're screwed from the start).

you've still to explain what the 9 you add to the right of all the other 9s in 0.9999... signifies, and to state why it must be true that since x is not y, a statement P(x) cannot be the same as statement Q(y), such as Hurkyl's 1,2 and 2,4 thing with 1/2=2/4.

why should my system agree with something that is wrong?

fractions are a completely different notation system, do that in decimals :| and none of that 9xInfinity=10xInfinity "omg 9 =10" crap. that's just stupid.

and Zurtex you have disproved NOTHING. Good work again.

matt grime

May16-04, 06:38 AM

let's stay away from opinion shall we.

in your claim to get closer to 1 than 0.99999... by 'adding a nine' after all the nines already there, what does that actually mean numerically?

I mean i can say that there is an integer between 1 and two, called derek, but it doesn't mean there actually is one there does it?

so what does it mean to add a nine after all the nines in 0.99....?

and you've still not answered any of the 1/2 2/4 thing. and who on earth says 9=10? all that that proves is infinity is not in the multipliactive group of real numbers, which is not surprising because, erm, it isn't there. after all 0*9=0*10, but that doesn't imply 9=10.

so what does the notation mean of adding that nine after all the infinite number of nines there? this is the sixth time i believe i've asked you and you're yet to provide an answer.

Zurtex

May16-04, 07:02 AM

and Zurtex you have disproved NOTHING. Good work again.

:approve: Once again you show your non understanding of mathematics, stop bugging people who give their help free and willingly and actually go and learn some maths.

Hurkyl

May16-04, 11:14 AM

why should my system agree with something that is wrong?

Are you going to tell me that you're not talking about the standard system here either?

arildno

May16-04, 01:00 PM

why should my system agree with something that is wrong?

fractions are a completely different notation system
Oh how interesting!
I never knew that!
(Here I've gone about believing that the decimal notation is simply a shorthand for sums of certain fractions, but now I know differently!)

ram2048

May16-04, 07:18 PM

and you've still not answered any of the 1/2 2/4 thing. and who on earth says 9=10? all that that proves is infinity is not in the multipliactive group of real numbers, which is not surprising because, erm, it isn't there. after all 0*9=0*10, but that doesn't imply 9=10.

YOU PEOPLE are saying 9 = 10. in order for .999~ to be = to 1, there MUST exist a digit that is 10. since we're using base 10, a DIGIT of 10 cannot exist so that means you're wrong from the start.

so what does the notation mean of adding that nine after all the infinite number of nines there? this is the sixth time i believe i've asked you and you're yet to provide an answer.

Set infinity(d) as cardinal f(n) n=0 for the system would be a way of describing what i'm doing :|

and you had to ask 6 times because i told you already even if i described it to you, you'd just say "that's stupid, our system doesn't work like that, you're wrong"

ram2048

May16-04, 07:19 PM

and Zurtex you're just a useless bandwagoner, i hope to god that i never unwittingly convince you, because i don't want you "on my side" ;D

ram2048

May16-04, 07:30 PM

(Here I've gone about believing that the decimal notation is simply a shorthand for sums of certain fractions, but now I know differently!)

quite true.

decimals represent an exact quantity, and create a TRUE number line from which EXACTLY ACCURATE quantified values of things can be plotted.

fractions can only represent rational numbers on their own, or all as sums of fractions (exTREMEly tedious, which is why we use decimals), but fractions are also unique in that their quantity is defined as a ratio of two integers. thusly in THAT notation you can have 1/2 = 2/4 but the notation also has its own rules.

in decimal notation system ONE NUMBER will always be equal to ONLY itself.

PS] if you say "OMG you're wrong 1.3 = 01.30 = 001.300~" i will find you and cause you bodily harm

ram2048

May16-04, 07:34 PM

why should my system agree with something that is wrong?
Are you going to tell me that you're not talking about the standard system here either?

no i'm not going to tell you "that" because i have no idea what you're hinting at ... :P

Root(2) is making me angry. gonna go buy a sci-calc later this week see if i'm getting the right numbers.

Hurkyl

May16-04, 08:16 PM

no i'm not going to tell you "that" because i have no idea what you're hinting at ... :P

So you're unwilling to say something unless you can be sure I can't turn it around on you? Not too confindent are you? :tongue:

"that's stupid, our system doesn't work like that, you're wrong"

He's right, ya know. Your arguments that the standard system is "wrong" are based entirely on how your system works.

Root(2) is making me angry. gonna go buy a sci-calc later this week see if i'm getting the right numbers.

Windows calculator has a scientific mode!

ram2048

May16-04, 08:34 PM

well when i've had 18 or so pages of people trying desperately to turn stuff against me the caution comes naturally. it's not so much fear, as the tedium it takes to go back and correct my spelling or explain what i mean or change a term i'm using because it has "mathematical" meaning and i'm not using it right :O

He's right, ya know. Your arguments that the standard system is "wrong" are based entirely on how your system works

because my system is accurate and logical. i don't have .999~ = 1. I don't have infinity + 1 = infinity. I don't have 1/infinity = 0. I don't have infinity/infinity = undefined or infinity - infinity = undefined. I don't have infinity as a limit to impede your archemedial assertations. i have ways to obtain meaningful values using "infinitessimals" and other numbers that possess characteristics of infinity.

but most importantly is the system adheres to logic

ram2048

May16-04, 08:35 PM

Windows calculator has a scientific mode!

by JOVE it freaking does.

omg omg <thanks>

hello3719

May16-04, 09:51 PM

YOU PEOPLE are saying 9 = 10. in order for .999~ to be = to 1, there MUST exist a digit that is 10. since we're using base 10, a DIGIT of 10 cannot exist so that means you're wrong from the start.

what do you mean by a digit that is 10 ?
we can just have 9.999~ = 10 if you meant number
and for
1 * 0 = 0
2*0 = 0
...
will you also change the definition of zero in your system since you think that

infinity *1 = inifinity
infinity * 2 = infinity
...
is illogical?

ram2048

May16-04, 10:47 PM

i might...

0^0 = 1?

hmmm... i can't even picture that but i guess it's necessary for some calculations....

we'll see

Zurtex

May17-04, 03:38 AM

lol, now your threatening bodily harm because of your lack of understanding of mathematics. Basically you can even see flaws in your own argument but you seem unwilling to admit them to yourself. Anyway:

0^0 \neq 1

Can you think why?

Edit: I could be wrong on that actually, I would be interested to see if I am right.

I am fairly sure that 0^0 is undefined. But Microsoft calculator seems to disagree with me :confused:

matt grime

May17-04, 05:17 AM

come one ram, you could at least pretend to explain what it is that you think that the number 0.999... plus another 9 after all of those is. because at the moment you've not managed to provide one single example of what this real number allegedly is.

Note that decimals are not exact representatives of real numbers, that is not any of the definitions of real numbers that works, as you've noticed, now why don't you look up the words cauchy sequence equivalence and class, or pssoibly dedekind and cut.

point out one sigle example where anyone but you has stated that they think 9=10. it would appear you don't understand the counter arguments to your system.

you also seem to have not noticed that we've otld you about systems of infinitesimals where there are 'numbers' smaller than 1/n for all n, and ordinals where there are infinte ordinals such as w, with w and w+1 not equal. one presumes this is because you think you are talking about real numbers alone, so let's do some analysis, prove that x^2 is a continuous function in your system. you can't as it happens, since 1.n does'nt tend to zero in your system, or it can't if you'r being consistent.

Hurkyl

May17-04, 05:47 AM

In the real numbers, 0^0 is left undefined becuase no matter how you tried to define it, exponentiation would be discontinuous there. (proof based on the fact 0^x = 0 but x^0 = 1 for all positive x)

It tends to be convenient to adopt the convention that 0^0 = 1 for some applications.

Zurtex

May17-04, 05:57 AM

ram2048

May17-04, 10:57 AM

hello3719

May17-04, 12:49 PM

[QUOTE=ram2048
if you accept line Z) you prove my system works for digits beyond infinity. you're also accepting infinity-infinity=0.

QUOTE]
who said infinity - infnity can't be equal to 0 ? Did our system imply that?

ram2048

May17-04, 01:02 PM

your system doesn't define a way for infinity - infinity to be 0. not that it couldn't be, just that it was never looked at.

my system DOES define relations amongst equal and non-equal infinities.

but that's beside the point. in THIS case you'd be accepting equal infinities to cancel out. thusly you'd be proving my point about the existance of the digit 9 in position f(n) n=infinity+1

hello3719

May17-04, 01:38 PM

your system doesn't define a way for infinity - infinity to be 0. not that it couldn't be, just that it was never looked at.

in our system : lim as x->infinity of (x-x) is of type "infinity - infinity " and it does equal 0 in this case. Never looked at? very funny it seems that you didn't even take a calculus course.You don't even understand our system that's why you are trying to build a new one.

in THIS case you'd be accepting equal infinities to cancel out. thusly you'd be proving my point about the existance of the digit 9 in position f(n) n=infinity+1

IF A implies B, it doesn't mean that B implies A, your logic is flawed because of this. So we wouldn't be proving anything about your nonsense.

matt grime

May17-04, 01:59 PM

i keep hounduing you about the nines because you've not explained what on earth you mean by 'adding a 9' after the infinitely many that are there, this is the same as your intent that we ''create" a zero at the end of some string after multiplication by 10.

look back at your post 191 where you say you want to add another 9 after all of them to get a number even closer to 1, but we won't "allow" it.

matt grime

May17-04, 02:08 PM

Matt why do you keep hounding me about the 9?

my system works perfectly. in fact i have used your accepted conventions to prove my system.

consider x=.999~ where x possesses infinite number of digits 9.

Z) 10x - x = 9.999~ - .999~ = 9 in your system.

see:

.999~ infinite 9's
9.999~ infinite 9's +1

this is PROVEN TRUE if you accept the statement claimed in line Z) as a consequence of that calculation.

the logical conclusion to that is that in position f(n) where n=infinty+1 there existed a digit 9. (which would be false because we defined .999~ to have exactly infinite 9's but whatever)

if you accept line Z) you prove my system works for digits beyond infinity. you're also accepting infinity-infinity=0.

and every time you say .999~ = 1 you're saying 9 = 10 since in the expression .999~ there only exists digits 9. to be equal to 1, one of those 9's MUST BE a 10 somewhere in that infinite number of 9's.

in base 10 you don't HAVE digits of 10. you have 0-9. so you're wrong to even assume there might be.

you are saying that infinity is "equal" to the "number" of digits after the decimal point in the expansion of .999...., so it is a cardinal, define the arithmetic of your cardinals then. (it can be done). for instance, take 0.99.... and 0.88888... they have both an infinite number of digits, you "infinity", now i interleave them 0.989898... so i must have added the infinities together!, yet it must also be true that there are the same number digits, thus 2*infinity=infinity. so where's the arithemetic wrong there.

now, can you even remotely rigorously prove that 0.999=1 implies 9=10 using the proper definitions of addition and such? i can't see where you've done that, in fact you haven't, but then you've never even begun to understand the definition of the REAL numbers.

ram2048

May17-04, 03:42 PM

you are saying that infinity is "equal" to the "number" of digits after the decimal point in the expansion of .999...., so it is a cardinal, define the arithmetic of your cardinals then. (it can be done). for instance, take 0.99.... and 0.88888... they have both an infinite number of digits, you "infinity", now i interleave them 0.989898... so i must have added the infinities together!, yet it must also be true that there are the same number digits, thus 2*infinity=infinity. so where's the arithemetic wrong there.

that's absolutely wrong.

were you to take .999~ with exactly infinite 9's and .888~ with exactly infinite 8's and "interleave" them to where f(-1) is 9 f(-2) is 8 f(-3) is 9 etc each time pulling from the available 9's and 8's given, you would come up with 2x(original infinity) number of digits, BUT the sum total of those digits does NOT equal (original infinity).

yet it must also be true that there are the same number digits

very very wrong. any given infinity is equal to itself not equal to other infinities unless you clearly define the relationship beforehand, as i have been doing with digits and integers.

every time you bring something else it just gives me more and more reasons why my system is superior to the current.

in THIS case you'd be accepting equal infinities to cancel out. thusly you'd be proving my point about the existance of the digit 9 in position f(n) n=infinity+1

IF A implies B, it doesn't mean that B implies A, your logic is flawed because of this. So we wouldn't be proving anything about your nonsense.

If you're accepting that a 9 DOES exist BEYOND the number of infinite digits set forth in the initial expression such that you can multiply by 10 and a new digit 9 is brought "into play" to make the cancelling of 9's ABSOLUTELY PERFECT, then you're ACCEPTING the existence of f(n) n=infinity+1

this is a flat out consequence of YOUR logic.

that's NOT what my system believes in, it's just my system's way of interpreting the actions of your system. If you have a better way of describing to me HOW an EXACTLY infinite number of digits 9 can be multiplied by 10 and another 9 is "created" to maintain the equality you're welcome to explain it to me.

matt grime

May17-04, 03:51 PM

so you are insisting that there is an infinite plus one spot in a decimal expansion which implies the existence of an infinite'th spot. none of those things exists.

we do not create any more nines who on earth except you thinks we do? your intuition is completely wrong. stick to things you can prove.

what does the 9 in this alleged infinite'th place signify? there is no such place, stop pretending that we think there is.

but the string .98989898.... must hve exactly 'infinity' digits in it - how can you dsitinguish between it and the one where i interleave two strings .9999.,.. and .888...?

so far all you've shown is that you do not understand mathematics and that you are incapable of defining a consistent notation.

hello3719

May17-04, 04:02 PM

that's NOT what my system believes in, it's just my system's way of interpreting the actions of your system. If you have a better way of describing to me HOW an EXACTLY infinite number of digits 9 can be multiplied by 10 and another 9 is "created" to maintain the equality you're welcome to explain it to me.

Well first of all our infinite isn't a real number if you didn't understand that yet. Saying "exactly infinite" isn't proper concerning our infinite.
Tell me, if i have a never-ending supply of apples and I eat one will i still have a never-ending supply of apples?

matt grime

May17-04, 04:37 PM

so we've ascertained that your infinity is a cardinal, which you insisted it wasn't.

please offer formal statements about when two cardinals are equal.

ram2048

May17-04, 05:05 PM

my infinity is NOT cardinal

my "default infinity" can be cardinal. it can be many things because it is a tool that i define at the onset of calculations and extrapolate meanings from that point onwards.

without definition "default infinity" has no meaning whatsoever.

Well first of all our infinite isn't a real number if you didn't understand that yet. Saying "exactly infinite" isn't proper concerning our infinite.
Tell me, if i have a never-ending supply of apples and I eat one will i still have a never-ending supply of apples?

well then explain your infinite apples "getting eaten" such that EXACTLY none are left in 999~ - .999~. By your accounting "a never ending supply of apples" should still be "a never ending supply of apples" regardless of eating "an infinite number of apples" from that amount.

so therefore:
.999~ - .999~ = .999~ (your logic)

it is clear that your infinity cannot handle such contradictions. i don't know why you cling to it so desperately.

Integral

May17-04, 05:15 PM

my infinity is NOT cardinal

my "default infinity" can be cardinal. it can be many things because it is a tool that i define at the onset of calculations and extrapolate meanings from that point onwards.

without definition "default infinity" has no meaning whatsoever.

well then explain your infinite apples "getting eaten" such that EXACTLY none are left in 999~ - .999~. By your accounting "a never ending supply of apples" should still be "a never ending supply of apples" regardless of eating "an infinite number of apples" from that amount.

so therefore:
.999~ - .999~ = .999~ (your logic)

it is clear that your infinity cannot handle such contradictions. i don't know why you cling to it so desperately.

Pure and simple nonsense.

.999... - .999... = 1-1=0

Apparently you are unable to differentiate between an infinite number of digits and a quanity of infinite magnitude. All real numbers have an infinite number of digits, no real number is infinite in magnitude.

hello3719

May17-04, 05:25 PM

yea it seems he thinks that .999.... is infinite LOL.

ram2048

May17-04, 05:47 PM

Integral

May17-04, 05:52 PM

Please repeat that post in meaningful terms. What are you saying? It does not make any sense to me.

Once again ALL real numbers have an infinite number of digits, NO real number is infinite in magnitude. Now what is your point.

ram2048

May17-04, 06:00 PM

you know perfectly well what i'm talking about.

feigning ignorance doesn't help your case one bit.

Integral

May17-04, 06:17 PM

Am I supposed to read you mind? Your post does not make any sense. It does not appear to address any of issues being discussed. How does it relate to your previous nonsense statement that 1-1=1 ?

For the third time .
All real numbers have an infinite number of digits NO real number is infinite in magnitude.

So yes, .999... being a real number has an infinite number of digits. What is the problem?

ram2048

May17-04, 07:11 PM

i'm just going to assume then that you've not read any of my posts concerning this matter, Integral, as i have explained several times in exacting detail WHAT i'm talking about.

how many digits of with value 9 does .999~ have?

i'm also assuming you KNOW what a digit is since you claim all "real numbers" have infinite of them

each digit position has a value from 0 to 9. simple question to you, how many digits have value of 9?

hello3719

May17-04, 07:25 PM

that's what .999~ means. infinite digits. meaning if i were to COUNT each digit as f(n) i would count to infinity (magnitude).

it's your number and your definition. is the number of digits NOT equal to infinity were they to be matched on a 1 to 1 basis?

if no then you would have to define exactly what ~ or ... or _ means in your statement.

and even supposing you DO define it, come back and explain to me how in a system where everything to the left of a number is greater than and everything to the right is less than, you have come up with a number to the right of another that is exactly equal to a number to the left.

I don't see a problem. When I write ~ or ... or _ we mean that the pattern repeats itself never-endingly. We can then say that .999... = .999... (obvious)
do you agree that .999... is between 0 and 2. If yes, knowing that 0 and 2 are real numbers then .999... must be a real number. lets call x this real number. Then the way real numbers are defined x = x implying x - x = 0
x=.999...
10x = 9.999... = 9 + .999... = 9 + x
10x - x = 9
9x = 9
x=1

even if .999... has an infinity of digits we can still do the 1 to 1 basis since .999... = .999... and it is a real number. there is then no contradiction
with the fact that infinity- infinity being undeterminate since the number .999... itself isn't infinity.
And even if there is an infinity of the digit "9" it doesn't mean that infinity - infinity is ALWAYS equal to 0, in this case it is in some other cases it isn't our system is totally consistent for this matter.In your system you say that infinity - infinity is AlWAYS = 0 , the possibility of being 0 in our system is existant but not exclusive to it. Try to understand the uses of limits with infinity and everything will be clear.

ram2048

May17-04, 07:33 PM

you don't understand my system AT ALL.

my system allows for infinities to be equal, greater than, or less than each other. complete with a way to differentiate between such expressions of infinity so that there is no confusion.

with your system, maybe it's 0 maybe it's undeterminable or undefined or maybe it's equal. you never know because you use the term "infinity" as an all-inclusive term for so many things.

hello3719

May17-04, 07:47 PM

didn't your system say that infinity(d) - infinity(d) = 0 ?
it is always equal to 0 in your system isn't that right ?
then logically your other "infinities"(ridiculous) are built on infinity(d).

and about our system i said that infinity - infinity is AN UNDETERMINATE FORM not INFINITY. As i said a thousand times try to understand the avantages of undeterminate forms. Tell me, do you know how to use limits, find derivatives etc...
If you do you wouldn't see the undeterminate forms as an inconvenience but as a logical consequence and totally treatable form with regards to the concept.

ram2048

May17-04, 08:01 PM

hello3719 - and about our system i said that infinity - infinity is AN UNDETERMINATE FORM not INFINITY.

Well first of all our infinite isn't a real number if you didn't understand that yet. Saying "exactly infinite" isn't proper concerning our infinite.
Tell me, if i have a never-ending supply of apples and I eat one will i still have a never-ending supply of apples?

i made the comparison that .999~ was infinite un ending number of 9's just like in the case with your apples. you never replied back when i said that if they're infinite un-ending i should be able to eat an infinite amount of them without affecting that property. such that:

.999~ - .999~ = .999~ (unending) (minus) (infinite) (equals) (unending)

please verify.

didn't your system say that infinity(d) - infinity(d) = 0 ?
it is always equal to 0 in your system isn't that right ?
then logically your other "infinities"(ridiculous) are built on infinity(d).

i said that's how my system is built AT THE VERY BEGINNING. OMG you finally understand :P

Integral

May17-04, 09:09 PM

Ram,
Let's get one thing straight. You do NOT have a system. You have nothing of any use to anybody. Since you do not care to learn what generations of mathematicians have developed why should anybody care what nonsense you have cooked up in the last couple of hours?

Of course I do not understand your system, you have northing but misconceptions, why should I waste my time making any effort to understand nonsense?

You are the loser here, you have an opportunity to learn something useful but instead continue to insist that everyone but you is wrong. What a waste of time.

hello3719

May17-04, 10:29 PM

i made the comparison that .999~ was infinite un ending number of 9's just like in the case with your apples. you never replied back when i said that if they're infinite un-ending i should be able to eat an infinite amount of them without affecting that property. such that:

.999~ - .999~ = .999~ (unending) (minus) (infinite) (equals) (unending)

please verify.

Did you read my last post well? It seems you didn't, i explained you that .999... is a REAL NUMBER so .999... - .999... = 0
Re-read well my last post since you did not understand it in depth.

.999~ - .999~ = .999~ (unending) (minus) (infinite) (equals) (unending)

.999... is infinite?
i explained the number of digits thing in the last post

eating apples non-endingly from a non-ending supply of apples
can only be represented as infinite - infinite wich is undeterminate
you cannot say that it is equivalent to .999... - .999...

and when i talked about the apples i meant to tell you that you can take a finite number of apples of a non-ending supply but there will always be a non-ending supply of apples.
This was to show you the true nature of infinite.
I think you should have understood it by now.

Hurkyl

May17-04, 10:42 PM

i would count to infinity (magnitude).

But never arrive at infinity.

ram2048

May17-04, 11:59 PM

hello3719
I don't see a problem. When I write ~ or ... or _ we mean that the pattern repeats itself never-endingly

i explained you that .999... is a REAL NUMBER so .999... - .999... = 0
Re-read well my last post since you did not understand it in depth.

eating apples non-endingly from a non-ending supply of apples
can only be represented as infinite - infinite wich is undeterminate

.999~ - .999~ = indeterminate then?

oh apples can't be 9's? why can't they be. numbers are just tools for describing reality. if i substitute an apple for every 9 there's no failure in logic, just the contradiction you happily supplied

ram2048

May18-04, 12:03 AM

Integral: you don't want to read about it then you're free to leave and never look back.

fact of the matter is you're so scared i might be right that you're unwilling to apply any effort at all to understand me.

my "system" irons out alot of kinks in the current system and replaces a lot of false notions and assumptions with perfect logical ones.

you wouldn't lash out like this if you weren't feeling threatened by me.

ram2048

May18-04, 12:05 AM

But never arrive at infinity.

true enough

matt grime

May18-04, 03:58 AM

ram, when we say there are an infinite number of nines in the expansion 0.999... we mean exactly what the word literally means: that the number of them is not finite. that is all. we also can go further and say they are in bijection with N, tha natural numbers, by place.

we are not using infinity as a number in the same sense as a real number. that is why we have cardinals, and ordinals, which are dsitinct, and have different arithmetics. there are also infinitesimals. we also have analysis.

so you have developed a symbol, call ik K, that indicates the 'number' of 9s in the expression 0.99999....

how many elements are there in N or Z? how many finite groups are there?

seems like you're going to have to have a different one for every object unless you give a way of comparing them. is there a comparison?

Integral

May18-04, 04:23 AM

What is it that I should fear? You have not presented a single coherent concept. you have repeatably demonstrated your lack of knowledge or understanding of mathematics. you simply do not have the tools to formulate a meaningful mathematical statement. Your ideas are not to be feared they are to be laughed at.

Before you can even think about fixing something you must understand how it works. You do not understand the Real Number system, therefore have no hope of "fixing" it.

lets do a bit of simple arithmetic.

.999... - .999... =

( \Sigma_{n=1}^{\infty} 9* 10^{-n}) -( \Sigma_{n=1}^{\infty} 9* 10^{-n}) =
9 * ( \Sigma_{n=1}^{\infty} 10^{-n}) -9*( \Sigma_{n=1}^{\infty}10^{-n}) =
(9-9)* (\Sigma_{n=1}^{\infty} 10^{-n}) = 0

Are you able to comprehend simple arithmetic? Now why don't you find some other misconception to share with us? Clearly 1-1=0 there is no doubt, except in your system which seems to lead you to this result. I strongly object to you attempting to tell us about the results of a system you do not understand. I will not argue the results of YOUR "system" since it is nonsense I expect nothing from it. I will continue to correct you when you misrepresent the results of standard Real Analysis.

Hurkyl

May18-04, 06:33 AM

fact of the matter is you're so scared i might be right that you're unwilling to apply any effort at all to understand me.

Is that why you don't spend effort to understand the standard mathematical ideas?

numbers are just tools for describing reality.

No. Numbers are what is defined by mathematical definitions and/or axioms. Whether numbers are capable of describing reality is another question all together.

my "system" irons out alot of kinks

Since "kink" here means "disagrees with ram's intuition", this justification doesn't particularly motivate me.

replaces a lot of false notions and assumptions with perfect logical ones.

Actually, it seems the other way around to me. Through the rigorous application of logic, I can get from the axioms to any of the statements I've made about the real numbers.

Whereas all of your arguments are simply your intuition (which the rest of us obviously don't share). You've axiomized your system enough to prove that ∞(d) - ∞(d) = 0, and that ∞(d+1) - ∞(d) = 1... (in particular, I think those were the only axioms you presented) but you haven't even said what d is or can be... and this is a far cry from the things you are asserting about your numbers.

matt grime

May18-04, 07:42 AM

to echo hurkyl, all you#'ve done is state that you have a symbol that you claim is infinity, though you don't explain what it represents, or does beyond claiming it is the 'number of nines' in 0.99... (which is thus a cardinal, though why you won't accept that is a mystery)., that you can manipulate like a real number, that is you've defined an extension R[k] by adjoining k (picking a letter at random), an indeterminate, and clamining that it k is 'infinity' without explaining what that means. in what sense is k infinity, and what does the arithmetic of it mean.

hello3719

May18-04, 10:27 AM

.999~ - .999~ = indeterminate then?

oh apples can't be 9's? why can't they be. numbers are just tools for describing reality. if i substitute an apple for every 9 there's no failure in logic, just the contradiction you happily supplied

k then substitute each 9 for an apple.

i see where your intuition is leading, you mean to say to me that

0.infinity - 0.infinity is indeterminate wich is true in this form

( i used ridiculous notations only to satisfy your intuition)

do you know what indeterminate means? It means that we can only extrapolate the real value by means of other informations. the form in itself doesn't give ENOUGH INFORMATION BUT CAN ADMIT ANY NUMBER. The TRUE SOLUTION depends on the concept. When replacing each 9 by an apple we TOOK OUT INFORMATION

here we know that

0.infinity is a REAL NUMBER since it is between 0 and 2 , right?
We also know that we replaced .999... with 0.infnity
then IN THIS CASE the form 0.infinity - 0.infinity = 0 by properties of real numbers and the fact both represent the same number.
every number can be a solution to an indeterminate form so there is no contradiction in our definitions.
The information "REAL NUMBER" will eliminate the problem of indetermination in our present case and will give us THE ONLY CORRECT ANSWER wich is 0.

You know that you can even call 4 - 3 indeterminate
but when you use the fact that both are real numbers and use their main properties you will obviously say that 4-3 = 1 ,so being indeterminate isn't at all a problem

when we say that infinity - infinity is indeterminate it is because it can yield any number and + or - infinity.
if we take the variable x and y representing numbers , we can also say that x - y is indeterminate since we dont have enough information to solve it, but is solvable if you have both values of x and y. same idea with the treatment of infinity. Hope it makes things more clear for your intuition.